ONLINE PAGE PROOFS. Exponential functions Kick off with CAS 11.2 Indices as exponents

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1 . Kick off with CAS. Indices as eponents Eponential functions.3 Indices as logarithms.4 Graphs of eponential functions.5 Applications of eponential functions.6 Inverses of eponential functions.7 Review

2 . Kick off with CAS Eponential functions Using CAS technolog, sketch the following eponential functions on the same set of aes. a = b = 3 c = 5 d = 8 e = Using CAS technolog, enter = a into the function entr line and use a slider to change the value of a. 3 When sketching an eponential function, what is the effect of changing the value of a in the equation? 4 Using CAS technolog, sketch the following eponential functions on the same set of aes. a = b = + c = + 5 d = 3 e = 5 5 Using CAS technolog, enter = + k into the function entr line and use a slider to change the value of k. 6 When sketching an eponential function, what is the effect of changing the value of k in the equation? 7 Using CAS technolog, sketch the following eponential functions on the same set of aes. a = b = c = +3 d = 5 e = 8 8 Using CAS technolog, enter = h into the function entr line and use a slider to change the value of h. 9 When sketching an eponential function, what is the effect of changing the value of h in the equation? 0 On the one set of aes, sketch the graphs of = and = +. Describe the transformations to get from to. Please refer to the Resources tab in the Prelims section of our ebookplus for a comprehensive step-b-step guide on how to use our CAS technolog.

3 . Units & AOS Topic 8 Concept Indices as eponents Concept summar Practice questions Indices as eponents Inde or eponential form When the number 8 is epressed as a power of, it is written as 8 = 3. In this form, the base is and the power (also known as inde or eponent) is 3. The form of 8 epressed as 3 is known both as its inde form with inde 3 and base, and its eponential form with eponent 3 and base. The words inde and eponent are interchangeable. For an positive number n where n = a, the statement n = a is called an inde or eponential statement. For our stud: the base is a where a R + \ the eponent, or inde, is where R the number n is positive, so a R +. Inde laws control the simplification of epressions which have the same base. Review of inde laws Recall the basic inde laws: From these, it follows that: Fractional indices Since a in inde form as 3 a m a n = a m+n a m a n = a m n (a m ) n = a mn a 0 = a n = a and n a (ab) n = a n b n a b a b n = an b n n = b a n = an = a and ( a) = a, then a = a. Thus, surds such as 3 can be written. The smbols, 3,... n are radical signs. An radical can be converted to and from a fractional inde using the inde law: a n = n a n n n A combination of inde laws allows am to be epressed as am = a can also be epressed as am n = a m n = n a m. n m = ( n a) m. It am n = n a m or am n = n a m 57 Maths Quest MATHEMATICAL METHODS VCE Units and

4 WORKeD eample a Epress 3n 9 n as a power of 3. 8 n 5 b Simplif 3a b (a b ). c Evaluate think WritE a Epress each term with the same base. a 3 n 9 n = 3n (3 ) n 8 n (3 4 ) n Appl an appropriate inde law. Note: To raise a power to a power multipl the indices. 3 Appl an appropriate inde law. Note: To multipl numbers with the same base, add the indices. 4 Appl an appropriate inde law and state the answer. Note: To divide numbers with the same base, subtract the indices. = 3n 3 ( n) 3 4(n ) = 3n 3 n 3 4n 4 = 3n+ n 3 4n 4 = 3 3 4n 4 = 3 (4n 4) = 3 6 4n 3n 9 n = 3 6 4n 8 n b Use an inde law to remove the brackets. b 5 3a b (a b ) = 3 a 4 b 5 a b 4 Appl the inde laws to terms with the same base to simplif the epression. c Epress each term with a positive inde. Note: There is no inde law for the addition of numbers. Appl the inde law for fractional indices. Note: The fractional indices could be interpreted in other was including, for eample, as 8 3 = ( 3 ) 3 Evaluate each term separatel and calculate the answer required. 3. = 9a 4 b 5 a b 4 = 8a 6 b = 8a 6 b c = 8 = = () + 3 = 4 + = 5 topic eponential FunCtIOns 573

5 WORKeD eample Indicial equations An indicial equation has the unknown variable as an eponent. In this section we shall consider indicial equations which have rational solutions. Method of equating indices If inde laws can be used to epress both sides of an equation as single powers of the same base, then this allows indices to be equated. For eample, if an equation can be simplified to the form 3 = 4, then for the equalit to hold, 3 = 4. Solving this linear equation gives the solution to the indicial equation as = 4 3. In the case of an inequation, a similar method is used. If 3 < 4, for eample, then 3 < 4. Solving this linear inequation gives the solution to the indicial inequation as < 4. 3 To solve for the eponent in equations of the form a = n: Epress both sides as powers of the same base. Equate the indices and solve the equation formed to obtain the solution to the indicial equation. Inequations are solved in a similar manner. However, ou need to ensure the base a is greater than prior to epressing the indices with the corresponding order sign between them. For eample, ou first need to write < 3 as < 3 and then solve the linear inequation < 3 to obtain the solution > 3. Solve = for. 5 think Use the inde laws to epress the left-hand side of the equation as a power of a single base. Epress the right-hand side as a power of the same base. WritE = Indicial equations which reduce to quadratic form The technique of substitution to form a quadratic equation ma be applicable to indicial equations (4 ) = = 5 8 = = = 3 Equate indices and calculate the required value of. Equating indices, 8 = 3 = Maths Quest MatheMatICaL MethODs VCe units and

6 WORKeD eample 3 think To solve equations of the form p a + q a + r = 0: Note that a = (a ). Reduce the indicial equation to quadratic form b using a substitution for a. Solve the quadratic and then substitute back for a. Since a must alwas be positive, solutions for can onl be obtained for a > 0; reject an negative or zero values for a. Solve = 0 for. Use a substitution technique to reduce the indicial equation to quadratic form. Note: The subtraction signs prevent the use of inde laws to epress the left-hand side as a power of a single base. WritE = 0 Let a = 3 a 6a 7 = 0 Solve the quadratic equation. (a 9)(a + 3) = 0 a = 9, a = 3 3 Substitute back and solve for. Replace a b 3. 3 = 9 or 3 = 3 (reject negative value) 3 = 9 3 = 3 = scientific notation (standard form) Inde notation provides a convenient wa to epress numbers which are either ver large or ver small. Writing a number as a 0 b (the product of a number a where a < 0 and a power of 0) is known as writing the number in scientific notation (or standard form). The age of the earth since the Big Bang is estimated to be ears, while the mass of a carbon atom is approimatel grams. These numbers are written in scientific notation. To convert scientific notation back to a basic numeral: move the decimal point b places to the right if the power of 0 has a positive inde, in order to obtain the large number a 0 b represents; or move the decimal point b places to the left if the power of 0 has a negative inde, in order to obtain the small number a 0 b represents. This is because multipling b 0 b is equivalent to dividing b 0 b. signifi cant fi gures When a number is epressed in scientific notation as either a 0 b or a 0 b, the number of digits in a determines the number of significant figures in the basic numeral. The age of the Earth is ears in scientific notation or ears to three significant figures. To one significant figure, the age would be ears. topic eponential FunCtIOns 575

7 WORKeD eample 4 a Epress each of the following numerals in scientific notation and state the number of significant figures each numeral contains. i ii b Epress the following as basic numerals. i ii think a i Write the given number as a value between and 0 multiplied b a power of 0. Note: The number is large so the power of 0 should be positive. b Count the number of digits in the number a in the scientific notation form a 0 b and state the number of significant figures. ii Write the given number as a value between and 0 multiplied b a power of 0. Note: The number is small so the power of 0 should be negative. Count the number of digits in the number a in the scientific notation form and state the number of significant figures. i Perform the multiplication. Note: The power of 0 is positive, so a large number should be obtained. ii Perform the multiplication. Note: The power of 0 is negative, so a small number should be obtained. EErCisE. PraCtisE Work without Cas Indices as eponents WE a Epress n 8 +n b Simplif (9a 3 b 4 ) a c Evaluate n as a power of.. Simplif 0p5 m 3 q 5(p q 3 ) 4m. b. 3 WE Solve = for. 4 Solve the following inequations. a < 75 WritE a i In scientific notation, = There are 5 significant figures in the number ii = has 6 significant figures. b i Move the decimal point 9 places to the right = ii Move the decimal point 5 places to the left = b 9 3 > Maths Quest MatheMatICaL MethODs VCe units and

8 Consolidate Appl the most appropriate mathematical processes and tools 5 WE3 Solve = 0 for. 6 Solve 48 = 3 for. 7 WE4 a Epress each of the following numbers in scientific notation, and state the number of significant figures each number contains. i ii b Epress the following as basic numerals. i ii Calculate (4 0 6 ) (5 0 3 ) without using a calculator. 9 a Epress the following in inde (eponent) form. i a 3 b 4 ii b 3 a b 4 b Epress the following in surd form. 3 i a b ii iii Evaluate without a calculator: 3 a 4 b c (6 (3 ) 5 5 ) d 5 0 Simplif and epress the answer with positive indices. 5 a 5 3 b 3 a 3( ) 3 b a 3 (ab) (3 4 ) 3a 3 b ( 8a ) b c (mn ) 0n4 m m 3 n 3(m n) d 3 4m n m n ( 3m 3 n ) e m n m n f 4 (4 ) 3 43 a Epress as a power of. 6 b Epress 3+n 8 n as a power of n c Epress (0.) 3 as a power of 0. d Epress 5n+ 5 n as a power of Solve for : 5 a = 4 b c 9 7 = 3 d 3 3 > 7 8 e = 8 f = Topic Eponential functions 577

9 Master 4 Use a suitable substitution to solve the following equations. a = 0 b = 3 c = 0 d ( + ) = 4 e 0 0 = 99 f = 0 5 a Epress in scientific notation: i ii the diameter of the Earth, given its radius is 6370 km iii iv the distance between Roland Garros and Kooong tennis stadiums of km. b Epress as a basic numeral: i ii c Epress the following to significant figures: i people attended a football match. ii The probabilit of winning a competition is iii The solution to an equation is = iv The distance flown per ear b the Roal Fling Doctor Service is km. 6 If = , show that 3 3 = 0 7 a Solve the pair of simultaneous equations for and : 8 If 5 = = 0.0 b Solve the pair of simultaneous equations for a and k. a k = 40 a k = 0 3a n 3 a n+ = , determine the values of the constants a and n. 9 Evaluate: a and epress the answer in scientific notation to 4 significant figures b.9.3e to 4 significant figures and eplain what the.3e notation means c (3 0 9 ), epressing the answer in standard form d ( ), epressing the answer in standard form to 4 significant figures. 0 a Simplif. 3 5 b Solve the equations: i 5 5 = to obtain eactl 5 ii 5 5 = 0.5 to obtain to 4 significant figures. 578 Maths Quest MATHEMATICAL METHODS VCE Units and

10 .3 Units & AOS Topic 8 Concept Indices as logarithms Concept summar Practice questions Indices as logarithms Not all solutions to indicial equations are rational. In order to obtain the solution to an equation such as = 5, we need to learn about logarithms. Inde-logarithm forms A logarithm is also another name for an inde. The inde statement, n = a, with base a and inde, can be epressed with the inde as the subject. This is called the logarithm statement and is written as = log a (n). The statement is read as equals the log to base a of n (adopting the abbreviation of log for logarithm). The statements n = a and = log a (n) are equivalent. n = a = log a (n), where the base a R + \, the number n R + and the logarithm, or inde, R. Consider again the equation = 5. The solution is obtained b converting this inde statement to the logarithm statement. = 5 = log (5) The number log (5) is irrational: the power of which gives the number 5 is not rational. A decimal approimation for this logarithm can be obtained using a calculator. The eact solution to the indicial equation = 5 is = log (5); an approimate solution is.39 to 5 significant figures. Not all epressions containing logarithms are irrational. Solving the equation = 8 b converting to logarithm form gives: = 8 = log (8) In this case, the epression log (8) can be simplified. As the power of which gives the number 8 is 3, the solution to the equation is = 3; that is, log (8) = 3. Use of a calculator Calculators have two inbuilt logarithmic functions. Base 0 logarithms are obtained from the LOG ke. Thus log 0 () is evaluated as log(), giving the value of to 4 decimal places. Base 0 logarithms are called common logarithms and, in a previous era, the were commonl used to perform calculations from tables of values. The are also known as Briggsian logarithms in deference to the English mathematician Henr Briggs who first published their table of values in the seventeenth centur. Base e logarithms are obtained from the LN ke. Thus log e () is evaluated as ln (), giving the value to 4 decimal places. Base e logarithms are called natural or Naperian logarithms, after their inventor John Napier, a seventeenth-centur Scottish baron with mathematical interests. These logarithms occur etensivel in calculus. The number e itself is known as Euler s number and, like π, it is a transcendental irrational number that has great importance in higher mathematical studies, as ou will begin to discover in Units 3 and 4 of Mathematical Methods. CAS technolog enables logarithms to bases other than 0 or e to be evaluated. Topic Eponential functions 579

11 WORKeD eample 5 a Epress 3 4 = 8 as a logarithm statement. b Epress log 7 = as an inde statement. 49 c Solve the equation 0 =.8, epressing the eponent to significant figures. d Solve the equation log 5 () = for. think WritE a Identif the given form. a 3 4 = 8 The inde form is given with base 3, inde or logarithm 4 and number 8. Convert to the equivalent form. Since n = a = log a (n), 8 = = log 3 (8). The logarithm statement is 4 = log 3 (8). b Identif the given form. b log 7 49 = The logarithm form is given with base 7, number and logarithm, or inde, of. 49 Convert to the equivalent form. Since = log a (n) n = a = log = 7 The inde statement is 49 = 7. c Convert to the equivalent form. c 0 =.8 = log 0 (.8) Evaluate using a calculator and state the. to significant figures answer to the required accurac. Note: The base is 0 so use the LOG ke on the calculator. d Convert to the equivalent form. d log 5 () = = 5 Calculate the answer. = 5 Logarithm laws Since logarithms are indices, unsurprisingl there is a set of laws which control simplification of logarithmic epressions with the same base. For an a, m, n > 0, a, the laws are:. log a () = 0. log a (a) = 3. log a (m) + log a (n) = log a (mn) 4. log a (m) log a (n) = log a m n 5. log a (m p ) = plog a (m). Note that there is no logarithm law for either the product or quotient of logarithms or for epressions such as log a (m ± n). 580 Maths Quest MatheMatICaL MethODs VCe units and

12 Proofs of the logarithm laws. Consider the inde statement. a 0 = log a () = 0. Consider the inde statement. a = a log a (a) = 3. Let = log a (m) and = log a (n). m = a and n = a mn = a a Convert to logarithm form: Substitute back for and : 4. With and as given in law 3: = a + + = log a (mn) log a (m) + log a (n) = log a (mn) m n = a = a Converting to logarithm form, and then substituting back for and gives: Substitute back for and : 5. With as given in law 3: a = log a m n log a (m) log a (n) = log a = log a (m) m = a Raise both sides to the power p: (m) p = (a ) p m p = a p Epress as a logarithm statement with base a: p = log a (m p ) Substitute back for : p log a (m) = log a (m p ) m n Topic Eponential functions 58

13 WORKeD eample 6 Simplif the following using the logarithm laws. a log 0 (5) + log 0 () b log (80) log (5) c log 3 (a 4 ) + log 3 a d log a 4 log a () think WritE a Appl the appropriate logarithm law. a log 0 (5) + log 0 () = log 0 (5 ) = log 0 (0) Simplif and state the answer. = since log a (a) = b Appl the appropriate logarithm law. b log (80) log (5) = log 80 5 = log (6) Further simplif the logarithmic epression. = log ( 4 ) = 4log () 3 Calculate the answer. = 4 = 4 c Appl a logarithm law to the second term. Note: As with indices, there is often more than one wa to approach the simplification of logarithms. Appl an appropriate logarithm law to combine the two terms as one logarithmic epression. c log 3 (a 4 ) + log 3 a = log (a 4 a ) + log a = log (a 4 ) + log a = log (a 4 a ) + log a = log 3 a 4 a = log 3 (a 5 ) 3 State the answer. = 5log 3 (a) d Epress the numbers in the logarithm terms in inde form. Note: There is no law for division of logarithms. Simplif the numerator and denominator separatel. 3 Cancel the common factor in the numerator and denominator. d log a 4 log a () = log a ( ) log a () = log a() log a () = log a() log a () = a 58 Maths Quest MatheMatICaL MethODs VCe units and

14 WORKeD eample 7 Logarithms as operators Just as both sides of an equation ma be raised to a power and the equalit still holds, taking logarithms of both sides of an equation maintains the equalit. If m = n, then it is true that log a (m) = log a (n) and vice versa, provided the same base is used for the logarithms of each side. This application of logarithms can provide an important tool when solving indicial equations. Consider again the equation = 5 where the solution was given as = log (5). Take base 0 logarithms of both sides of this equation. = 5 log 0 ( ) = log 0 (5) Using one of the logarithm laws, this becomes log 0 () = log 0 (5) from which the solution to the indicial equation is obtained as = log 0(5). This form of the solution log 0 () can be evaluated on a scientific calculator and is the prime reason for choosing base 0 logarithms in solving the indicial equation. It also demonstrates that log (5) = log 0(5), which is a particular eample of another log 0 () logarithm law called the change of base law. Change of base law for calculator use The equation a = p for which = log a (p) could be solved in a similar wa to = 5, giving the solution as = log 0(p) log 0 (a). Thus log a(p) = log 0(p). This form enables log 0 (a) decimal approimations to logarithms to be calculated on scientific calculators. The change of base law is the more general statement allowing base a logarithms to be epressed in terms of an other base b as log a (p) = log b(p). This more general log b (a) form shall be left until Units 3 and 4. Convention There is a convention that if the base of a logarithm is not stated, this implies it is base 0. As it is on a calculator, log(n) represents log 0 (n). When working with base 0 logarithms it can be convenient to adopt this convention. a State the eact solution to 5 = 8 and calculate its value to 3 decimal places. b Calculate the eact value and the value to 3 decimal places of the solution to the equation = 6. think a Convert to the equivalent form and state the eact solution. WritE a 5 = 8 = log 5 (8) The eact solution is = log 5 (8). topic eponential FunCtIOns 583

15 Use the change of base law to epress the answer in terms of base 0 logarithms. Since log a (p) = log 0(p) log 0 (a) then log 5 (8) = log 0(8) log 0 (5) = log 0(8) log 0 (5) 3 Calculate the approimate value..9 to 3 decimal places. b Take base 0 logarithms of both sides. Note: The convention is not to write the base 0. Appl the logarithm law so that terms are no longer eponents. 3 Solve the linear equation in. Note: This is no different to solving an other linear equation of the form a b = c ecept the constants a, b, c are epressed as logarithms. 4 Calculate the approimate value. Note: Remember to place brackets around the denominator for the division. b = 6 Take logarithms to base 0 of both sides: log( ) = log(6 ) ( ) log() = log(6) Epand: log() log() = log(6) Collect terms together: log() = log(6) + log() = (log(6) + log()) log() = log(6) + log() This is the eact solution to 3 decimal places. equations containing logarithms While the emphasis in this topic is on eponential (indicial) relations for which some knowledge of logarithms is essential, it is important to know that logarithms contribute substantiall to Mathematics. As such, some equations involving logarithms are included, allowing further consolidation of the laws which logarithms must satisf. Remembering the requirement that must be positive for log a () to be real, it is advisable to check an solution to an equation involving logarithms. An value of which when substituted back into the original equation creates a log a (negative number) term must be rejected as a solution. Otherwise, normal algebraic approaches together with logarithm laws are the techniques for solving such equations. WORKeD eample 8 Solve the equation log 6 () + log 6 ( ) = for. think Appl the logarithm law which reduces the equation to one logarithm term. WritE log 6 () + log 6 ( ) = log 6 (( )) = log 6 ( ) = 584 Maths Quest MatheMatICaL MethODs VCe units and

16 Convert the logarithm form to its equivalent form. Note: An alternative method is to write log 6 ( ) = log 6 (6) from which = 6 is obtained. Eercise.3 PRactise Work without CAS Q 4 Indices as logarithms WE5 a Epress 5 4 = 65 as a logarithm statement. b Epress log 36 (6) = as an inde statement. c Solve the equation 0 = 8.5, epressing the eponent to significant figures. d Solve the equation log 3 () = for. a Evaluate log e (5) to 4 significant figures and write the equivalent inde statement. b Evaluate to 4 significant figures and write the equivalent logarithm statement. 3 WE6 Simplif using the logarithm laws: a log (3) + log (4) b log (9) log () 3 c log 3 (3a 3 ) log 3 a d log a(8) log a (4) 4 Given log a () = 0.3 and log a (5) = 0.7, evaluate: a log a (0.5) b log a (.5) c log a (0) 5 WE7 a State the eact solution to 7 = 5 and calculate its value to 3 decimal places. b Calculate the eact value and the value to 3 decimal places of the solution to the equation 3 +5 = 4. 6 If log (3) log () = log () + log (5), solve for. 7 WE8 Solve the equation log 3 () + log 3 ( + ) = for. 8 Solve the equation log 6 () log 6 ( ) = for. Converting from logarithm form to inde form gives: = 6 = 6 3 Solve the quadratic equation. 6 = 0 ( 3)( + ) = 0 = 3, = 4 Check the validit of both solutions in the original equation. 5 State the answer. The solution is = 3. Check in log 6 () + log 6 ( ) = If = 3, LHS = log 6 (3) + log 6 () = log 6 (6) = = RHS If =, LHS = log 6 ( ) + log 6 ( 3) which is not admissible. Therefore reject =. Topic Eponential functions 585

17 Consolidate Appl the most appropriate mathematical processes and tools 9 a Epress as a logarithm statement with the inde as the subject: 3 i 5 = 3 ii 4 = 8 iii 0 3 = 0.00 b Epress as an inde statement: i log (6) = 4 ii log 9 (3) = iii log 0 (0.) = 0 Rewrite each of the following in the equivalent inde or logarithm form and hence calculate the value of. a = log 8 b log 5 () = 0.5 c 0 () = 4. Epress the answer to decimal places. d 3 = e. Epress the answer to decimal places. e log (5) = 3 f log (5) = Use the logarithm laws to evaluate the following. a log 9 (3) + log 9 (7) b log 9 (3) log 9 (7) c log (4) + log (6) log () d log 5 ( log 3 (3)) 7 e log 3 + log f log 3 3 log 3 7 log 9 Simplif the following. log a (9) a b log 4 (3 n + 3 n+ ) log a ( 5 3) c log 8 (6) log 6 (8) d log 0 (000) + log a Epress log (0) in terms of log 0 (). b State the eact solution and then give the approimate solution to 4 significant figures for each of the following indicial equations. i = 8 ii 5 = 8 iii 7 = 3 c Obtain the approimate solution to 4 significant figures for each of the following inequations. i 3 0 ii 5 > 0.4 d Solve the following equations. i log5() = 8 ii log() = 7 4 Solve the indicial equations to obtain the value of to decimal places. a 7 = 4 b 0 = 5 c 5 9 = 3 7 d = 6 3 e = f = 5 5 Solve the following for. a log ( + ) + log ( ) = 3log (3) b log 3 () + log 3 (4) = log 3 ( + ) log 3 () c log ( + ) log (3) = 4 d log () + log ( ) = e (log 0 () + 3)(log 4 () 3) = 0 f log 3 () = log 3 ( 3) 586 Maths Quest MATHEMATICAL METHODS VCE Units and

18 Master.4 Units & AOS Topic 8 Concept 3 Graphs of eponential functions Concept summar Practice questions Interactivit Eponential functions int Given log a (3) = p and log a (5) = q, epress the following in terms of p and q. a log a (5) b log a (5) c log a (45) d log a (0.6) e log a Epress in terms of. f log a ( 5) log a ( 7) a log 0 () = log 0 () + b log ( ) = c log e log 0 (0 3 ) = 3 = 6 d = 0 f 0 3log 0() = 8 Solve the following equations, giving eact solutions. a = 0 b 5 5 = 4 c = 0 d log a ( 3 ) + log a ( ) 4log a () = log a () e (log ()) log ( ) = 8 f log 0 ( 3 ) log 0 ( + ) = log 0() 9 a Give the solution to = 50 to 4 significant figures. b Give the eact solution to the equation log(5) + log( + 5) =. 0 a Evaluate log 0 (5) + log 5 (0) with the calculator on Standard mode and eplain the answer obtained. b Evaluate log () log () and eplain how the result is obtained. Graphs of eponential functions Eponential functions are functions of the form f : R R, f() = a, a R + \. The provide mathematical models of eponential growth and eponential deca situations such as population increase and radioactive deca respectivel. The graph of = a where a > Before sketching such a graph, consider the table of values for the function with rule = = From the table it is evident that > 0 for all values of, and that as, 0. This means that the graph will have a horizontal asmptote with equation = 0. It is also evident that as, with the values increasing rapidl. = (, ) (0, ) = 0 0 Topic Eponential functions 587

19 Since these observations are true for an function = a where a >, the graph of = will be tpical of the basic graph of an eponential with base larger than. Ke features of the graph of = and an such function = a where a > : horizontal asmptote with equation = 0 -intercept is (0, ) shape is of eponential growth domain R range R + one-to-one correspondence For =, the graph contains the point (, ); for the graph of = a, a >, the graph contains the point (, a), showing that as the base increases, the graph becomes steeper more quickl for values > 0. This is illustrated b the graphs of = and = 0, with the larger base giving the steeper graph. The graph of = a where 0 < a < An eample of a function whose rule is in the form = a where 0 < a < is =. Since =, the rule for the graph of this eponential function = where the base lies between 0 and is identical to the rule = where the base is greater than. The graph of = shown is tpical of the graph of = a where a > and of the graph of = a where 0 < a <. Ke features of the graph of = and an such function with rule epressed as either = a where 0 < a < or as = a where a > : horizontal asmptote with equation = 0 -intercept is (0, ) shape is of eponential deca domain R range R + one-to-one correspondence reflection of = in the -ais The basic shape of an eponential function is either one of growth or deca. (, 0) = (, ) (0, ) = 0 (, ) (0, ) = 0 0 ( ), 0 = 0 = (0, ) (0, ) (0, ) = 0 = 0 = 0 = a, a > = a, a > = a, 0 < a < As with other functions, the graph of = a will be inverted (reflected in the -ais). 588 Maths Quest MATHEMATICAL METHODS VCE Units and

20 WORKeD eample 9 a On the same set of aes, sketch the graphs of = 5 and = 5, stating their ranges. b Give a possible equation for the graph shown. think WritE a Identif the asmptote of the a = 5 first function. The asmptote is the line with equation = 0. Find the -intercept. -intercept: when = 0, = (0, ) 3 Calculate the coordinates of a Let =. second point. = 5 = 5 (, 5) 4 Use the relationship between the = 5 two functions to deduce the ke This is the reflection of = 5 in the -ais. features of the second function. The graph of = 5 has the same asmptote as that of = 5. Equation of its asmptote is = 0. Its -intercept is (0, ). Point (, 5) lies on the graph. 5 Sketch and label each graph. (, 5) (0, ) = 0 0 (0, ) = 5 = 5 (, 5) 6 State the range of each graph. The range of = 5 is R + and the range of = 5 is R. b Use the shape of the graph to b The graph has a deca shape. suggest a possible form for the rule. Let the equation be = a. Use a given point on the graph to calculate a. The point (, 5) 5 = a a = 5 3 State the equation of the graph. The equation of the graph could be = 5. The equation could also be epressed as = = = 0 or topic eponential FunCtIOns 589

21 WORKeD eample 0 think translations of eponential graphs Once the basic eponential growth or eponential deca shapes are known, the graphs of eponential functions can be translated in similar was to graphs of an other functions previousl studied. the graph of = a + k Under a vertical translation the position of the asmptote will be altered to = k. If k < 0, the graph will have -ais intercepts which are found b solving the eponential equation a + k = 0. the graph of = a h Under a horizontal translation the asmptote is unaffected. The point on the -ais will no longer occur at =. An additional point to the -intercept that can be helpful to locate is the one where = h, since a h will equal when = h. A horizontal translation and a vertical translation of the graph of = 3 are illustrated in the diagram b the graphs of = 3 + and = 3 respectivel. Under the horizontal translation of unit to the left, the point (0, ) (, ); under the vertical translation of unit down, the point (, 3) (, ). (, ) = 3 + = 3 (0, 3) (, 3) (, ) (0, ) = 3 0 = Sketch the graphs of each of the following and state the range of each. a = 4 a State the equation of the asmptote. WritE Calculate the -intercept. -intercept: let = 0, = 4 = 3 -intercept is (0, 3). 3 Calculate the -intercept. -intercept: let = 0, 4 = 0 = 4 = b = 0 (+) a = 4 The vertical translation 4 units down affects the asmptote. The asmptote has the equation = 4. = -intercept is (, 0). 590 Maths Quest MatheMatICaL MethODs VCe units and

22 4 Sketch the graph and state the range. A growth shape is epected since the coefficient of is positive. = 4 (, 0) 0 b Identif the ke features from the given equation. Calculate the coordinates of a second point on the graph. 3 Sketch the graph and state the range. (0, 3) Range is ( 4, ). = 4 b = 0 (+) Reflection in -ais, horizontal translation unit to the left. The asmptote will not be affected. Asmptote: = 0 There is no -intercept. -intercept: let = 0, = 0 = 0 -intercept is (0, 0.). Let = = 0 0 = The point (, ) lies on the graph. A deca shape is epected since the coefficient of is negative. = 0 (+) (, ) (0, 0.) 0 = 0 Range is R +. Dilations Eponential functions of the form = b a have been dilated b a factor b (b > 0) from the -ais. This affects the -intercept, but the asmptote remains at = 0. Topic Eponential functions 59

23 WORKeD eample think Eponential functions of the form = a n have been dilated b a factor (n > 0) from the -ais. This n affects the steepness of the graph but does not affect either the -intercept or the asmptote. A dilation from the -ais of factor and a dilation from the -ais of factor of the graph of = 3 are illustrated in the diagram b the graphs of = 3 and = 3 respectivel. Under the dilation from the -ais of factor, the point (, 3) (, 6); under the dilation from the -ais of factor, the point (, 3), 3. Combinations of transformations Eponential functions with equations of the form = b a n( h) + k are derived from the basic graph of = a b appling a combination of transformations. The ke features to identif in order to sketch the graphs of such eponential functions are: the asmptote the -intercept the -intercept, if there is one. Another point that can be obtained simpl could provide assurance about the shape. Alwas aim to show at least two points on the graph. Sketch the graphs of each of the following and state the range of each. a = 0 5 a Identif the ke features using the given equation. Calculate the coordinates of a second point. WritE a = 0 5 asmptote: = 0 no -intercept -intercept: let = 0 = 0 5 = 0 5 = -intercept is (0, ). b = 3 Since the horizontal translation is to the right, let =. = 0 5 = = 0 = 0 Point, 0 lies on the graph. (, 6) = 3 = 3 = 3 (, 3) (0, ) (, (0, ) 3) 0 59 Maths Quest MatheMatICaL MethODs VCe units and

24 3 Sketch the graph and state the range. (0.5, 0) = 0 5 (0, ) b Write the equation in the form = b a n( h) + k and state the asmptote. 0 Range is R +. b = 3 = 3 + Asmptote: = Calculate the -intercept. -intercept: let = 0 = = 3 + = -intercept is (0, ). 3 Calculate the -intercept. Note: As the point (0, ) lies below the asmptote and the graph must approach the asmptote, there will be an -intercept. 4 Calculate an approimate value for the -intercept to help determine its position on the graph. -intercept: let = 0 0 = 3 = 0 = 3 In logarithm form, = log 3 = log (3 ) = log (3) = log (3) The eact -intercept is (log (3), 0). = log (3) = log 0(3) log 0 ().58 The -intercept is approimatel (.58, 0). Topic Eponential functions 593

25 5 Sketch the graph and state the range. Note: Label the -intercept with its eact coordinates once the graph is drawn. = 0 (log (3), 0) (0, ) = 3 Eercise.4 PRactise Work without CAS Graphs of eponential functions WE9 a On the same set of aes, sketch the graphs of = 3 and = 3, stating their ranges. b Give a possible equation for the graph shown. Sketch the graphs of = (.5) and = 3 on the same set of aes. 3 WE0 Sketch the graphs of each of the following and state the range of each. a = 4 b = 3 (+) 4 Sketch the graph of = 4 + and state its range. 5 WE Sketch the graphs of each of the following and state the range of each. a = 0 Range is (, ). (, 3) b = = 0 6 The graph shown has the equation = a.3 + b. Determine the values of a and b. (0, 0) 0 = (0, ) 594 Maths Quest MATHEMATICAL METHODS VCE Units and

26 Consolidate Appl the most appropriate mathematical processes and tools 7 a i Sketch, on the same set of aes, the graphs of = 4, = 6 and = 8. ii Describe the effect produced b increasing the base. b i Sketch, on the same set of aes, the graphs of =, = and = ii Epress each rule in a different form. 8 a i Sketch, on the same set of aes, the graphs of = 5, = 7 and = 9. ii Describe the effect produced b increasing the base. b i Sketch, on the same set of aes, the graphs of = (0.8), = (.5) and = (0.8). ii Describe the relationships between the three graphs. 9 Sketch each of the following graphs, showing the asmptote and labelling an intersections with the coordinate aes with their eact coordinates. a = 5 + b = 4 c = 0 d = 6.5 (.5) 0 Sketch the graphs of: a = b = 3 + c = d = 7 Sketch the graphs of: 3 a = 3 4 b = c = 3 3 d =.5 0 a Sketch the graphs of = 3 and = 9 and eplain the result. b i Use inde laws to obtain another form of the rule for = ii Hence or otherwise, sketch the graph of = a Determine a possible rule for the given graph in the form = a.0 + b. 0 (0, 5) = 3 b The graph of an eponential function of the form = a.3 k contains the points (, 36) and (0, 4). Determine its rule and state the equation of its asmptote. c For the graph shown, determine a possible rule in the form = a 3 b. d Epress the equation given in part c in another form not involving a horizontal = 6 translation. (0, 0). Topic Eponential functions 595

27 Master.5 Units & AOS Topic 8 Concept 4 Applications of eponential functions Concept summar Practice questions 4 Sketch the graphs of the following eponential functions and state their ranges. Where appropriate, an intersections with the coordinate aes should be given to decimal place. a = 0 0 b = 5 c = 3 3 d = (3.5) + 7 e = f = Consider the function f : R R, f() = 3 6. a Evaluate: i f() ii f(0), epressing the answer in simplest surd form. b For what value of, if an, does: i f() = 9 ii f() = 0 iii f() = 9? c Sketch the graph of = f() and state its range. d Solve the inequation f() correct to significant figures. 6 Use a graphical means to determine the number of intersections between: a = and =, specifing an interval in which the -coordinate of an point of intersection lies. b = and = c = e and = d = + and = sin () e = 3 and = 6, determining the coordinates of an points of intersection algebraicall. f = and = 6, giving the coordinates of an points of intersection. 7 Obtain the coordinates of the points of intersection of = and =. 8 Sketch the graphs of = 33 () and = 33 () + and compare their asmptotes, - and -intercepts and the value of their -coordinates when = 0. What transformation maps to? Applications of eponential functions The importance of eponential functions lies in the frequenc with which the occur in models of phenomena involving growth and deca situations, in chemical and phsical laws of nature and in higher-level mathematical analsis. Eponential growth and deca models For time t, the eponential function defined b = b a nt where a > represents eponential growth over time if n > 0 and eponential deca over time if n < 0. The domain of this function would be restricted according to the wa the independent time variable t is defined. The rule = b a nt ma also be written as = b.a nt. In some mathematical models such as population growth, the initial population ma be represented b a smbol such as N 0. For an eponential deca model, the time it takes for 50% of the initial amount of the substance to deca is called its half-life. 596 Maths Quest MATHEMATICAL METHODS VCE Units and

28 WORKeD eample think a Calculate the initial amount. The deca of a radioactive substance is modelled b Q(t) = Q 0.7 kt where Q kg is the amount of the substance present at time t ears and Q 0 and k are positive constants. a Show that the constant Q 0 represents the initial amount of the substance. b If the half-life of the radioactive substance is 00 ears, calculate k to one significant figure. c If initiall there was 5 kg of the radioactive substance, how man kilograms would deca in 0 ears? Use the value of k from part b in the calculations. b Form an equation in k from the given information. Note: It does not matter that the value of Q 0 is unknown since the Q 0 terms cancel. Solve the eponential equation to obtain k to the required accurac. c Use the values of the constants to state the actual rule for the eponential deca model. Calculate the amount of the substance present at the time given. WritE a Q(t) = Q 0.7 kt The initial amount is the value of Q when t = 0. Let t = 0: Q(0) = Q = Q 0 Therefore Q 0 represents the initial amount of the substance. b The half-life is the time it takes for 50% of the initial amount of the substance to deca. Since the half-life is 00 ears, when t = 00, Q(00) = 50% of Q 0 Q(00) = 0.50Q 0... () From the equation, Q(t) = Q 0.7 kt. When t = 00, Q(00) = Q 0.7 k(00) Q(00) = Q k... () Equate equations () and (): 0.50Q 0 = Q k Cancel Q 0 from each side: 0.50 =.7 00k Convert to the equivalent logarithm form. 00k = log.7 (0.5) k = 00 log.7(0.5) = 00 log 0(0.5) log 0 (.7) c Q 0 = 5, k = Q(t) = t When t = 0, Q(0) = topic eponential FunCtIOns 597

29 3 Calculate the amount that has decaed. Note: Using a greater accurac for the value of k would give a slightl different answer for the amount decaed. Since =.68, in 0 ears approimatel.68 kg will have decaed. WORKeD eample 3 analsing data One method for detecting if data has an eponential relationship can be carried out using logarithms. If the data is suspected of following an eponential rule such as = A 0 k, then the graph of log() against should be linear. The reasoning for this is as follows: = A 0 k A = 0k log A = k log() log(a) = k log() = k + log(a) This equation can be written in the form Y = k + c where Y = log() and c = log(a). The graph of Y versus is a straight line with gradient k and vertical ais Y-intercept (0, log(a)). Such an analsis is called a semi-log plot. While eperimental data is unlikel to give a perfect fit, the equation would describe the line of best fit for the data. Logarithms can also be effective in determining a power law that connects variables. If the law connecting the variables is of the form = p then log() = p log(). Plotting log() values against log() values will give a straight line of gradient p if the data does follow such a law. Such an analsis is called a log-log plot. For a set of data {(, )}, plotting log() versus log() gave the straight line shown in the diagram. Form the equation of the graph and hence determine the rule connecting and. log() 0 (0, 3) (, 0) log() think State the gradient and the coordinates of the intercept with the vertical ais. WritE Gradient = rise run = 3 Intercept with vertical ais: (0, 3) 598 Maths Quest MatheMatICaL MethODs VCe units and

30 Form the equation of the line. Let Y = log() and X = log(). The equation of the line is Y = mx + c where m = 3, c = 3. 3 Epress the equation in terms of the variables marked on the aes of the given graph. 4 Collect the terms involving logarithms together and simplif to create a logarithm statement. 5 Epress the equation with as the subject. Note: Remember the base of the logarithm is 0. Eercise.5 PRactise Applications of eponential functions Therefore the equation of the line is Y = 3 X + 3. The vertical ais is log() and the horizontal ais is log(), so the equation of the graph is log() = 3 log() + 3. log() =.5 log() + 3 log() +.5 log () = 3 log() + log(.5 ) = 3 log(.5 ) = 3 log 0 (.5 ) = 3.5 = 0 3 WE The deca of a radioactive substance is modelled b Q(t) = Q 0.7 kt where Q is the amount of the substance present at time t ears and Q 0 and k are positive constants. a Show that the constant Q 0 represents the initial amount of the substance. b If the half-life of the radioactive substance is 300 ears, calculate k to one significant figure. c If initiall there was 50 kg of the radioactive substance, how man kilograms would deca in 0 ears? Use the value of k from part b in the calculations. The manager of a small business is concerned about the amount of time she spends dealing with the growing number of s she receives. The manager starts keeping records and finds the average number of s received per da t = can be modelled b D = 4 where D is the average number of s received per da t weeks from the start of the records. a How man dail s on average was the manager receiving when she commenced her records? log() b After how man weeks does the model predict that the average number of s received per da will double? (0, ) 3 WE3 For a set of data (, ), plotting log() versus log() gave the straight line shown in the diagram. ( 0.8, 0) 0 From the equation of the graph and hence determine the rule connecting and. log() Topic Eponential functions 599

31 Consolidate Appl the most appropriate mathematical processes and tools 4 For a set of data (, ), the semi-log plot of log() log() versus gave the straight line shown in the diagram. Form the equation of the graph and hence determine an eponential rule connecting (, 0.3) and. 0 (0, 0) 5 The value V of a new car depreciates so that its value after t ears is given b V = V 0 kt. a If 50% of the purchase value is lost in 5 ears, calculate k. b How long does it take for the car to lose 75% of its purchase value? 6 The number of drosophilae (fruit flies), N, in a colon after t das of observation is modelled b N = t. Give whole-number answers to the following. a How man drosophilae were present when the colon was initiall observed? b How man of the insects were present after 5 das? c How man das does it take the population number to double from its initial value? d Sketch a graph of N versus t to show how the population changes. e After how man das will the population first eceed 00? 7 The value of an investment which earns compound interest can be calculated from nt the formula A = P + r where P is the initial investment, r the interest rate n per annum (earl), n the number of times per ear the interest is compounded and t the number of ears of the investment. An investor deposits $000 in an account where interest is compounded monthl. a If the interest rate is 3% per annum: i Show that the formula giving the value of the investment is A = 000(.005) t. ii Calculate how much the investment is worth after a 6-month period. iii What time period would be needed for the value of the investment to reach $500? b The investor would like the $000 to grow to $500 in a shorter time period. What would the interest rate, still compounded monthl, need to be for the goal to be achieved in 4 ears? 8 A cup of coffee is left to cool on a kitchen table inside a Brisbane home. The temperature of the coffee T ( C) after t minutes is thought to be given b T = t. a B how man degrees does the coffee cool in 0 minutes? b How long does it take for the coffee to cool to 65 C? c Sketch a graph of the temperature of the coffee for t [0, 40]. d B considering the temperature the model predicts the coffee will eventuall cool to, eplain wh the model is not realistic in the long term. 600 Maths Quest MATHEMATICAL METHODS VCE Units and

32 9 The contents of a meat pie immediatel after being heated in a microwave have a temperature of 95 C. The pie is removed from the microwave and left to cool. A model for the temperature of the pie as it cools is given b T = a 3 0.3t + 5 where T is the temperature after t minutes of cooling. a Calculate the value of a. b What is the temperature of the contents of the pie after being left to cool for minutes? c Determine how long, to the nearest minute, it will take for the contents of the meat pie to cool to 65 C. d Sketch the graph showing the temperature over time and state the temperature to which this model predicts the contents of the pie will eventuall cool if left unattended. 0 The barometric pressure P, measured in kilopascals, at height h above sea level, measured in kilometres, is given b P = P o 0 kh where P o and k are positive constants. The pressure at the top of Mount Everest is approimatel one third that of the pressure at sea level. a Given the height of Mount Everest is approimatel 8848 metres, calculate the value of k to significant figures. Use the value obtained for k for the remainder of question 6. b Mount Kilimanjaro has a height of approimatel 5895 metres. If the atmospheric pressure at its summit is approimatel kilopascals, calculate the value of P 0 to 3 decimal places. c Use the model to estimate the atmospheric pressure to decimal places at the summit of Mont Blanc, 480 metres, and of Mount Kosciuszko, 8 metres in height. d Draw a graph of the atmospheric pressure against height showing the readings for the four mountains from the above information. The common Indian mnah bird was introduced into Australia in order to control insects affecting market gardens in Melbourne. It is now considered to be Australia s most important pest problem. In 976, the species was introduced to an urban area in New South Wales. B 99 the area averaged 5 birds per square kilometre and b 994 the densit reached an average of 75 birds per square kilometre. Topic Eponential functions 60

33 A model for the increasing densit of the mnah bird population is thought to be D = D 0 0 kt where D is the average densit of the bird per square kilometre t ears after 976 and D 0 and k are constants. a Use the given information to set up a pair of simultaneous equations in D and t. b Solve these equations to show that k = 3 log(5) and D 0 = and hence that k 0.33 and D c A project was introduced in 996 to curb the growth in numbers of these birds. What does the model predict was the average densit of the mnah bird population at the time the project was introduced in the ear 996? Use k 0.33 and D and round the answer to the nearest whole number. d Sometime after the project is successfull implemented, a different model for the average densit of the bird population becomes applicable. This model is given b D = 30 0 t3 + b. Four ears later, the average densit is reduced to 40 birds per square kilometre. How much can the average densit epect to be reduced? Carbon dating enables estimates of the age of fossils of once living organisms to be ascertained b comparing the amount of the radioactive isotope carbon-4 remaining in the fossil with the normal amount present in the living entit, which can be assumed to remain constant during the organism s life. It is known that carbon-4 decas with a half-life of approimatel 5730 ears according to an eponential model of the form C = C o, where C is the amount of the isotope remaining in the fossil t ears after death and C o is the normal amount of the isotope that would have been present when the organism was alive. a Calculate the eact value of the positive constant k. b The bones of an animal are unearthed during digging eplorations b a mining compan. The bones are found to contain 83% of the normal amount of the isotope carbon-4. Estimate how old the bones are. 3 a Obtain the equation of the given linear graphs and hence determine the relationship between and. i The linear graph of log 0 () against i log() log 0 () is shown. ii The linear graph of log () against (, 0) is shown. log() b The acidit of a solution is due (0, ) to the presence of hdrogen ions. The concentration of these ions is measured b the ph scale calculated as ph = log ([H + ]) where [H + ] is the ii log () concentration of hdrogen ions. i The concentration of hdrogen ions in bleach is 0 3 (0, 0) per mole and in pure water the concentration is 0 7 (4, ) per mole. What are the ph readings for bleach and for pure water? kt 60 Maths Quest MATHEMATICAL METHODS VCE Units and

34 Master ii Lemon juice has a ph reading of and milk has a ph reading of 6. Use scientific notation to epress the concentration of hdrogen ions in each of lemon juice and milk and then write these concentrations as numerals. iii Solutions with ph smaller than 7 are acidic and those with ph greater than 7 are alkaline. Pure water is neutral. How much more acidic is lemon juice than milk? iv For each one unit of change in ph, eplain the effect on the concentration of hdrogen ions and acidit of a solution. 4 The data shown in the table gives the population of Australia, in millions, in ears since (ears since 960) (population in millions) log() a Complete the third row of the table b evaluating the log() values to decimal places. b Plot log() against and construct a straight line to fit the points. c Show that the equation of the line is approimatel Y = where Y = log(). d Use the equation of the line to show that the eponential rule between and is approimatel = e After how man ears did the population double the 960 population? f It is said that the population of Australia is likel to eceed 8 million b the ear 030. Does this model support this claim? 5 Eperimental data ielded the following table of values: a Enter the data as lists in the Statistics menu and obtain the rule connecting the data b selecting the following from the Calc menu: i Eponential Reg ii Logarithmic Reg b Graph the data on the calculator to confirm which rule better fits the data. 6 Following a fall from his bike, Stephan is feeling some shock but not, initiall, a great deal of pain. However, his doctor gives him an injection for relief from the pain that he will start to feel once the shock of the accident wears off. The amount of pain Stephan feels over the net 0 minutes is modelled b the function Topic Eponential functions 603

35 .6 Units & AOS Topic 8 Concept 5 Inverses of eponential functions Concept summar Practice questions P(t) = (00t + 6).7 t, where P is the measure of pain on a scale from 0 to 00 that Stephan feels t minutes after receiving the injection. a Give the measure of pain Stephan is feeling: i at the time the injection is administered ii 5 seconds later when his shock is wearing off but the injection has not reached its full effect. b Use technolog to draw the graph showing Stephan s pain level over the 0-minute interval and hence give, to decimal places: i the maimum measure of pain he feels ii the number of seconds it takes for the injection to start lowering his pain level iii his pain levels after 5 minutes and after 0 minutes have elapsed. c Over the 0-minute interval, when was the effectiveness of the injection greatest? d At the end of the 0 minutes, Stephan receives a second injection modelled b P(t) = (00(t 0) + a).7 (t 0), 0 t 0. i Determine the value of a. ii Sketch the pain measure over the time interval t [0, 0] and label the maimum points with their coordinates. Inverses of eponential functions The inverse of = a, a R + \{} The eponential function has a one-to-one correspondence so its inverse must also be a function. To form the inverse of = a, interchange the - and -coordinates. function: = a domain R, range R + inverse function: = a domain R +, range R = log a () Therefore, the inverse of an eponential function is a logarithmic function: = log a () and = a are the rules for a pair of inverse functions. These are transcendental functions, not algebraic functions. However, the can be treated similarl to the inverse pairs of algebraic functions previousl encountered. This means the graph of = log a () can be obtained b reflecting the graph of = a in the line =. The graph of = log a (), for a > The shape of the basic logarithmic graph with rule = log a (), a > is shown as the reflection in the line = of the eponential graph with rule = a, a >. The ke features of the graph of = log a () can be deduced from those of the eponential graph. (, a) = a (0, ) (, 0) (a, ) = = log a 604 Maths Quest MATHEMATICAL METHODS VCE Units and

36 WORKeD eample 4 = a = log a () horizontal asmptote with equation = 0 vertical asmptote with equation = 0 -intercept (, 0) -intercept (0, ) point (, a) lies on the graph point (a, ) lies on the graph range R + domain R + domain R range R one-to-one correspondence one-to-one correspondence Note that logarithmic growth is much slower than eponential growth and also note > 0, if > that, unlike a which is alwas positive, log a () = 0, if = < 0, if 0 < < The logarithmic function is formall written as f : R + R, f() = log a (). a Form the eponential rule for the inverse of = log () and hence deduce the graph of = log () from the graph of the eponential. b Given the points (, ), (, 4) and (3, 8) lie on the eponential graph in part a, eplain how these points can be used to illustrate the logarithm law log (m) + log (n) = log (mn). think a Form the rule for the inverse b interchanging coordinates and then make the subject of the rule. WritE a = log () Inverse: = log () = Sketch the eponential function. = Asmptote: = 0 -intercept: (0, ) second point: let =, = Point (, ) is on the graph. 0 (0, ) = (, ) = 0 topic eponential FunCtIOns 605

37 3 Reflect the eponential graph in the line = to form the required graph. = log () has: asmptote: = 0 -intercept: (, 0) second point: (, ) (, ) = = log () = (0, ) (, ) b State the coordinates of the corresponding points on the logarithm graph. State the - and -values for each of the points on the logarithmic graph. 3 Use the relationship between the -coordinates to illustrate the logarithm law. The inverse of eponential functions of the form = b a n( h) + k The rule for the inverse of = b a n( h) + k is calculated in the usual wa b interchanging - and -coordinates to give = b a n( h) + k. Epressing this equation as an inde statement: k = b a n( h) k b = a n( h) Converting to the equivalent logarithm form: b log a k b 0 = n( h) (, 0) Given the points (, ), (, 4) and (3, 8) lie on the eponential graph, the points (, ), (4, ) and (8, 3) lie on the graph of = log (). = log () point (, ): when =, = point (4, ): when = 4, = point (8, 3): when = 8, = 3 The sum of the -coordinates of the points on = log () when = and = 4 equals the -coordinate of the point on = log () when = 8, as + = 3. log () + log (4) = log (8) log () + log (4) = log ( 4) This illustrates the logarithm law log (m) + log (n) = log (mn) with m = and n = Maths Quest MATHEMATICAL METHODS VCE Units and

38 WORKeD eample 5 think a Determine the range of the given function. State the domain of the inverse. b Form the rule for the inverse function b interchanging - and -coordinates and rearranging the equation obtained. Note: Remember to epress the rule for the inverse function f with f () in place of as its subject. Rearranging to make the subject: n log a k b = h = n log a k b The inverse of an eponential function is a logarithmic function. Since the corresponding pair of graphs of these functions must be smmetric about the line =, this provides one approach for sketching the graph of an logarithmic function. In this section, the graphs of logarithmic functions are obtained b deduction using the previousl studied ke features of eponential functions under a sequence of transformations. Should either the eponential or the logarithmic graph intersect the line = then the other graph must also intersect that line at eactl the same point. Due to the transcendental nature of these functions, technolog is usuall required to obtain the coordinates of an such point of intersection. Consider the function f : R R, f () = a What is the domain of its inverse? b Form the rule for the inverse function and epress the inverse function as a mapping. c Sketch = f () and = f () on the same set of aes. WritE a f() = Asmptote: = 3 -intercept: (0, 8) This point lies above the asmptote, so the range of f is (3, ). The domain of the inverse function is the range of the given function. The domain of the inverse function is (3, ). b Let f() =. Function: = Inverse: = = 5 3 = 5 Converting to logarithm form: 3 = log 5 3 = log 5 The inverse function has the rule f 3 () = log. 5 + h topic eponential FunCtIOns 607

39 Write the inverse function as a mapping. c Sketch the graph of the eponential function f and use this to deduce the graph of the inverse function f. The inverse function has domain (3, ) and its rule is f 3 () = log. 5 Hence, as a mapping: f : (3, ) R, f 3 () = log 5 c f() = 5 + 3, f 3 () = log 5 The ke features of f determine the ke features of f. = f() = f () asmptote: = 3 asmptote: = 3 -intercept (0, 8) -intercept (8, 0) second point on = f(): let = = = 3 Point (, 3) is on = f() and the point (3, ) is on = f (). = 3 (, 3) = log 3 5 = = (0, 8) (8, 0) 0 (3, ) The graphs intersect on the line =. Relationships between the inverse pairs As the eponential and logarithmic functions are a pair of inverses, each undoes the effect of the other. From this it follows that: log a (a ) = and a log a() = The first of these statements could also be eplained using logarithm laws: log a (a ) = log a (a) = = = Maths Quest MATHEMATICAL METHODS VCE Units and

40 WORKeD eample 6 think The second statement can also be eplained from the inde-logarithm definition that a n = n = log a (). Replacing n b its logarithm form in the definition gives: a n = a log a() = a Simplif log ( 3 ) using the inverse relationship between eponentials and logarithms. b Evaluate 0 log0(5). a Use inde laws to simplif the product of powers given in the logarithm epression. WritE a log ( 3 ) Consider the product 3. 3 = ( ) 3 = 4 3 = (4 3) = Simplif the given epression. log ( 3 ) = log ( ) = b Appl a logarithm law to the term in b 0 log 0(5) the inde. = 0 log 0(5) Simplif the epression. = 0 log 0(5) = 5 Therefore, 0 log0(5) = 5. transformations of logarithmic graphs Knowledge of the transformations of graphs enables the graph of an logarithmic function to be obtained from the basic graph of = log a (). This provides an alternative to sketching the graph as the inverse of that of an eponential function. Further, given the logarithmic graph, the eponential graph could be obtained as the inverse of the logarithmic graph. The logarithmic graph under a combination of transformations will be studied in Units 3 and 4. In this section we shall consider the effect a single transformation has on the ke features of the graph of = log a (). Dilations Dilations from either coordinate ais are recognisable from the equation of the logarithmic function: for eample, = log a () and = log a would give the images when = log a () undergoes a dilation of factor from the -ais and from the -ais respectivel. The asmptote at = 0 would be unaffected b either dilation. The position of the -intercept is affected b the dilation from the -ais as (, 0) (, 0). The dilation from the -ais does not affect the -intercept. topic eponential FunCtIOns 609

41 Horizontal translation: the graph of = log a ( h) The vertical asmptote will alwas be affected b a horizontal translation and this affects the domain of the logarithmic function. Under a horizontal translation of h units to the right or left, the vertical asmptote at = 0 must move h units to the right or left respectivel. Hence, horizontall translating the graph of = log a () b h units to obtain the graph of = log a ( h) produces the following changes to the ke features: equation of asmptote: = 0 = h domain: : > 0 : > h -intercept: (, 0) ( + h, 0) These changes are illustrated in the diagram b the graph = log () = 3 = 0 of = log () and its image, = log ( 3), after a horizontal translation of 3 units to the right. (, ) The diagram shows that the (5, ) = log ( 3) domain of = log ( 3) is (3, ). Its range is unaffected b the horizontal translation and remains R. It is important to realise that the domain and the asmptote position can be calculated algebraicall, since we onl take logarithms of positive numbers. For eample, the domain of 0 (, 0) (4, 0) = log ( 3) can be calculated b solving the inequation 3 > 0 > 3. This means that the domain is (3, ) as the diagram shows. The equation of the asmptote of = log ( 3) can be calculated from the equation 3 = 0 = 3. The function defined b = log a (n + c) would have a vertical asmptote when n + c = 0 and its domain can be calculated b solving n + c > 0. Vertical translation: the graph of = log a () + k Under a vertical translation of k units, neither the domain nor the position of the asmptote alters = log () = 0 from that of = log a (). The translated graph will have an -intercept which can be obtained b (, ) (8, 0) solving the equation log a () + k = 0. 0 (, 0) The graph of = log () 3 is a vertical (, ) translation down b 3 units of the graph of = log (). Solving log () 3 = 0 gives = 3 = log () 3 so the graph cuts the -ais at = 8, as illustrated. Reflections: the graphs of = log a () and = log a ( ) The graph of = log a () is obtained b inverting the graph of = log a (); that is, b reflecting it in the -ais. The graph of = log a ( ) is obtained b reflecting the graph of = log a () in the -ais. For log a ( ) to be defined, > 0 so the graph has domain : < Maths Quest MATHEMATICAL METHODS VCE Units and

42 = log a ( ) = 0 = log a () (, 0) 0 (, 0) WORKeD eample 7 think The relative positions of the graphs of = log a (), = log a () and = log a ( ) are illustrated in the diagram. The vertical asmptote at = 0 is unaffected b either reflection. a Sketch the graph of = log ( + ) and state its domain. b Sketch the graph of = log 0 () + and state its domain. c The graph of the function for which f() = log (b ) is shown below. i Determine the value of b. ii State the domain and range of, and form the rule for, the inverse function. iii Sketch the graph of = f (). WritE a Identif the transformation involved. a = log ( + ) Horizontal translation units to the left Use the transformation to state the equation of the asmptote and the domain. The vertical line = 0 the vertical line = under the horizontal translation. The domain is : >. 3 Calculate an intercepts with the coordinate aes. Note: The domain indicates there will -intercept: when = 0, = log () = be an intercept with the -ais as well as the -ais. = log a () = (0, ) 0 (, 0) -intercept (0, ) -intercept: when = 0, log ( + ) = 0 + = 0 + = = -intercept (, 0) Check: the point (, 0) (, 0) under the horizontal translation. topic eponential FunCtIOns 6

43 4 Sketch the graph. = (, 0) = log ( + ) (0, ) 0 b Identif the transformation involved. b = log 0 () + Vertical translation of unit upwards State the equation of the asmptote and the domain. 3 Obtain an intercept with the coordinate aes. 4 Calculate the coordinates of a second point on the graph. 5 Sketch the graph. The vertical transformation does not affect either the position of the asmptote or the domain. Hence, the equation of the asmptote is = 0. The domain is R +. Since the domain is R + there is no -intercept. -intercept: when = 0, log 0 () + = 0 log 0 () = = 0 = or intercept is (0., 0). Point: let =. = log 0 () + = 0 + = The point (, ) lies on the graph. Check: the point (, 0) (, ) under the vertical translation. = log 0 () + (, ) = 0 0 (0., 0) 6 Maths Quest MATHEMATICAL METHODS VCE Units and

44 c i State the equation of the asmptote shown in the graph and use this to calculate the value of b. Note: The function rule can be rearranged to show the horizontal translation and a reflection in the -ais. f() = log (b ) = log ( ( b)) The horizontal translation determines the position of the asmptote. ii Give the domain and range of the inverse function. Form the rule for the inverse function b interchanging - and -coordinates and rearranging the equation obtained. iii Use the features of the logarithm graph to deduce the features of the eponential graph. Sketch the graph of = f (). c i f() = log (b ) From the diagram, the asmptote of the graph is =. From the function rule, the asmptote occurs when: b = 0 = b Hence, b =. ii The given function has domain (, ) and range R. Therefore the inverse function has domain R and range (, ). function f : = log ( ) inverse f : = log ( ) = = f () = iii The ke features of f give those for f. = f() = f () asmptote: = asmptote = -intercept (, 0) -intercept (0, ) -intercept (0, ) -intercept (, 0) = (0, ) = f () (, 0) 0 Topic Eponential functions 63

45 Eercise.6 PRactise Consolidate Appl the most appropriate mathematical processes and tools Inverses of eponential functions WE4 a Form the eponential rule for the inverse of = log 0 () and hence deduce the graph of = log 0 () from the graph of the eponential. b Given the points (, 0), (, 00) and (3, 000) lie on the eponential graph in part a, eplain how these points can be used to illustrate the m logarithm law log 0 (m) log 0 (n) = log 0 n. Sketch the graphs of = and its inverse and form the rule for this inverse. 3 WE5 Consider the function f : R R, f() = 4 3. a What is the domain of its inverse? b Form the rule for the inverse function and epress the inverse function as a mapping. c Sketch = f() and = f () on the same set of aes. 4 Consider the function defined b = 4 + log (). a Form the rule for the inverse function. b Sketch the graph of the inverse function and hence draw the graph of = 4 + log () on the same set of aes. c In how man places do the two graphs intersect? 5 WE6 a Simplif log 6 ( 9 ) using the inverse relationship between eponentials and logarithms. b Evaluate 3 log (0). 6 Simplif 5 log 5() log 5 (3). 7 WE7 a Sketch the graph of = log 0 ( ) and state its domain. b Sketch the graph of = log 5 () and state its domain. c The graph of the function for which f() = log ( + b) is shown. i Determine the value of b. ii State the domain and range of, and form the rule for, the inverse function. iii Sketch the graph of = f (). = b (0, 0) 0 = log ( + b) 8 Sketch the graph of the function f : R + R, f() = log 4 () b identifing the transformations involved. 9 a On the same aes, sketch the graphs of = 3 and = 5 together with their inverses. b State the rules for the inverses graphed in part a as logarithmic functions. c Describe the effect of increasing the base on a logarithm graph. d On a new set of aes, sketch = 3 + and draw its inverse. e Give the equation of the inverse of = 3 +. f Sketch the graphs of = 5 + and its inverse on the same set of aes and give the rule for the inverse. 64 Maths Quest MATHEMATICAL METHODS VCE Units and

46 0 Sketch the graphs of the following functions and their inverses and form the rule for the inverse. a = 3 b = 8 c = 4 d = e = 0 f = Given f : R R, f() = 8 3 : a Determine the domain and the range of the inverse function f. b Evaluate f(0). c At what point would the graph of f cut the -ais? d Obtain the rule for f and epress f as a mapping. Consider the function defined b = (.5). a For what value of does =? b For what value of does = 0? c Sketch the graph of = (.5) showing the ke features. d On the same set of aes sketch the graph of the inverse function. e Form the rule for the inverse. f Hence state the solution to the equation (.5) = log.5. 3 Given g: (, ) R, g() = 3log 5 ( + ): a State the range of g. b Evaluate g(0). c At what point would the graph of g cut the -ais? d Obtain the rule for g and epress g as a mapping. e Sketch the graph of = g (). f Use the graph in part e to deduce the graph of = g(). 4 a Evaluate: i 3 log 3(8) ii 0 log 0()+log 0 (3) iii 5 log 5() iv 6 b Simplif: i 3 log 3() ii 3log () log 6(5) iii log ( ) + log 3 (9 6 ) iv log The diagram shows the graph of the eponential function = a+b + c. The graph intersects the line = twice and cuts the -ais at, 0 and the -ais at (0, ). = a+b + c a Form the rule for the eponential function. b Form the rule for the inverse function. = c For the inverse, state the equation of its asmptote and the coordinates of the points where its graph (,0 would cut the - and -aes. 0 d Cop the diagram and sketch the graph of the inverse (0, ) on the same diagram. How man points of intersection = 4 of the inverse and the eponential graphs are there? e The point (log (3), k) lies on the eponential graph. Calculate the eact value of k. f Using the equation for the inverse function, verif that the point (k, log (3)) lies on the inverse graph. ( Topic Eponential functions 65

47 6 Sketch the graphs of the following transformations of the graph of = log a (), stating the domain and range, equation of the asmptote and an points of intersection with the coordinate aes. a = log 5 () b = log 5 ( ) c = log 0 () + d = log 3 ( + ) e = log 3 (4 ) f = log ( + 4) 7 a Describe the transformations which map = log () = log ( ). b Use a logarithm law to describe the vertical translation which maps = log () = log (). c Use the change of base law to epress = log () in terms of base 0 logarithms and hence describe the dilation which would map = log 0 () = log (). d Hick s Law arose from research into the time taken for a person to make a decision when faced with a number of possible choices. For n equall probable options, the law is epressed as t = blog (n + ) where t is the time taken to choose an option, b is a positive constant and n. Draw a sketch of the time against the number of choices and show that doubling the number of options does not double the time to make the choice between them. 8 a The graph of the function with equation = alog 7 (b) contains the points (, 0) and (4, 4). Determine its equation. b The graph of the function with equation = alog 3 () + b contains the points, 8 and (, 4). 3 i Determine its equation. = ii Obtain the coordinates of the point where the graph of the inverse function would cut the -ais. (.5, 0) c i For the graph illustrated 0 in the diagram, determine a possible equation in the form = alog ( b) + c. (0, ) ii Use the diagram to sketch the graph of the inverse and form the rule for the inverse. d Consider the functions f and g for which f() = log 3 (4 + 9) and g() = log 4 ( 0.). i Determine the maimal domain of each function. ii State the equations of the asmptotes of the graphs of = f() and = g(). iii Calculate the coordinates of the points of intersection of each of the graphs with the coordinate aes. iv Sketch the graphs of = f() and = g() on separate diagrams. 66 Maths Quest MATHEMATICAL METHODS VCE Units and

48 Master 9 Obtain the coordinates of an points of intersection of the graph of = 3 with its inverse. Epress the values to 4 significant figures, where appropriate. 0 Consider the two functions with rules = log ( + 4) and = log () + log (4). a i Should the graphs of = log ( + 4) and = log () + log (4) be the same graphs? Use CAS technolog to sketch the graphs of = log ( + 4) and = log () + log (4) to verif our answer. ii Give an values of for which the graphs have the same value and justif algebraicall. b Sketch the graph of = log 3 (), stating its domain, range and tpe of correspondence. c Sketch the graph of = log 3 ( ), stating its domain, range and tpe of correspondence. d The graphs in parts b and c are not identical. Eplain wh this does not contradict the logarithm law log a (m p ) = plog a (m). Topic Eponential functions 67

49 ONLINE ONLY.7 Review the Maths Quest review is available in a customisable format for ou to demonstrate our knowledge of this topic. the review contains: short-answer questions providing ou with the opportunit to demonstrate the skills ou have developed to efficientl answer questions without the use of CAS technolog Multiple-choice questions providing ou with the opportunit to practise answering questions using CAS technolog ONLINE ONLY Activities to access ebookplus activities, log on to Interactivities A comprehensive set of relevant interactivities to bring difficult mathematical concepts to life can be found in the Resources section of our ebookplus. Etended-response questions providing ou with the opportunit to practise eam-stle questions. a summar of the ke points covered in this topic is also available as a digital document. REVIEW QUESTIONS Download the Review questions document from the links found in the Resources section of our ebookplus. studon is an interactive and highl visual online tool that helps ou to clearl identif strengths and weaknesses prior to our eams. You can then confidentl target areas of greatest need, enabling ou to achieve our best results. Units & Eponential functions Sit topic test 68 Maths Quest MatheMatICaL MethODs VCe units and