SAMPLE. Exponential Functions and Logarithms

Objectives C H A P T E R 5 Eponential Functions and Logarithms To define and understand eponential functions. To sketch graphs of the various tpes of eponential functions. To understand the rules for manipulating eponential and logarithmic epressions. To solve eponential equations. To evaluate logarithmic epressions. To solve equations using logarithmic methods. To sketch graphs of functions of the form = log a and simple transformations of this. To understand and use a range of eponential models. To sketch graphs of eponential functions. To appl eponential functions to solving problems. The function f () = ka,where k is a non zero constant and the base a is a positive real number other than, is called an eponential function (or inde function). Consider the following eample of an eponential function. Assume that a particular biological organism reproduces b dividing ever minute. The following table shows the population, P, after n one-minute intervals (assuming that all organisms are still alive). n 0 3 4 5 6 n P 4 8 6 3 64 n Thus P defines a function which has the rule P = n,aneponential (or inde) function. 406

Chapter 5 Eponential Functions and Logarithms 407 5. Graphs of eponential functions Two tpes of graphs will be eamined. Graphs of = a, a > Eample Plot the graph of =, and eamine the table of values for 3 3. A calculator can be used. 3 0 3 = 0.5 0.5 0.5 4 8 8 6 4 3 0 3 = It can be observed that, as negative values of increasing magnitude are considered, the graph approaches the -ais from above. The -ais is said to be an asmptote. Asthe magnitude of negative -values becomes larger, takes values closer and closer to zero, but never reaches zero, i.e. the graph gets closer and closer to the -ais. This is written as, 0 from the positive side, or as, 0 +. The -ais intercept for the graph is (0, ). The range of the function is R +. Eample Plot the graph of = 0 and eamine the table of values for. A calculator can be used. 0.5 0 0.5 = 0 0. 0.36 3.6 0 The -ais is an asmptote, and the -ais intercept is (0, ). The -values increase as the -values increase. This rate of increase for = 0 is greater than that for = for a given value of. 0 8 6 4 0.5 0 0.5 = 0

408 Essential Mathematical Methods &CAS It is worth noting at this stage that, for a and b positive numbers greater than, there is a positive number k such that a k = b. This can be seen from the graphs of = and = 0 above. Using a calculator to solve k = 0 graphicall gives k = 3.398...Hence 0 = ( 3.398... ) = 3.398.... This means that the graph of = 0 can be obtained from the graph of = badilation of factor k = from the -ais. 3.398... This shows that all graphs of the form = a,where a >, are related to each other b dilations from the -ais. This will be discussed again later in the chapter. Graphs of = a,0<a < Eample 3 Plot the graph of = ( ) and eamine the table of values for 3 3. A calculator can be used. 3 0 3 = ( ) = 8 4 0.5 0.5 0.5 The -ais is an asmptote, and the -ais intercept is (0, ). For this graph the -values decrease as the -values increase. This is written as, 0 from the positive side, or as, 0 +. The range of the function is R +. 8 6 4 3 0 3 = = In general: 0 = a, a > For 0 < a <, = a is equivalent to = b,where b = a. 0 = a, 0 < a <

Chapter 5 Eponential Functions and Logarithms 409 The graph of = a is obtained from the graph of = a bareflection in the -ais. Thus, for eample, the graph of = ( ) is obtained from the graph of = bareflection in the -ais, and vice versa. If a =, the resulting graph is a horizontal line with equation =. Eample 4 Plot the graph of = on a CAS calculator and hence find: a the value of when =., correct to 3 decimal places b the value of when = 9. Using the TI-Nspire Plot the graph. a Use Graph Trace (b 5 ) and tpe. to go straight to the point with -coordinate.. When =., = 4.87, correct to 3 decimal places. b The values of for which = 9 can be found b plotting the graph of = 9onthe same screen and finding the Intersection Point(s) (b 63). When = 9, = 3.70, correct to 3 decimal places.

40 Essential Mathematical Methods &CAS Using the Casio ClassPad Plot the graph for 3 4. a Click in the graph window and select Analsis G-solve -calc and enter =.. = 4.87 when =. (correct to 3 decimal places). b Select Analsis G-solve -calc and enter = 9. When = 9, = 3.70 (correct to 3 decimal places). Note: An alternative method is to enter = 9 and use G-solve Intersect. Transformations of eponential functions Eample 5 Sketch the graphs of each of the following functions. Give equations of asmptotes and the -ais intercepts, and state the range of each of the functions. (-ais intercepts need not be given.) a f : R R, f () = + 3 b f : R R, f () = 3 + c f : R R, f () = 3 +

Chapter 5 Eponential Functions and Logarithms 4 a For the function f : R R, f () = + 3, the corresponding graph is obtained b transforming the graph of = batranslation of 3 units in the positive direction of the -ais. The asmptote of =, with equation = 0, is transformed to the asmptote with equation = 3 for the graph of f () = + 3. The asmptotic behaviour can be described as, 3 from the positive side, or as, 3 +. As f (0) = 0 + 3 = 4, the -ais intercept is 4. The range of the function f : R R, f () = + 3is(3, ). 4 0 = + 3 = = 3 b For the function f : R R, f () = 3 +, the corresponding graph is obtained b transforming the graph of = 3 badilation of factor from the -ais, followed b a translation of unit in the positive direction of the -ais. The asmptote of = 3, with equation = 0, is transformed to the asmptote = for the graph of f () = 3 +. The asmptotic behaviour is described as, from the positive side, or as, +. The -ais intercept is given b f (0) = 3 0 + = 3. The range of the function f : R R, f () = 3 + is (, ). 0 = 3 + 3 = 3 = c For the function f : R R, f () = 3 +, the corresponding graph is obtained b transforming the graph of = 3 b reflection in the -ais followed b a translation of units in the positive direction of the -ais. The asmptote of = 3, with equation = 0, is transformed to the asmptote = for the graph of f () = 3 +.

4 Essential Mathematical Methods &CAS Eample 6 The asmptotic behaviour is described b as, from the negative side, or as,. The -ais intercept is given b f (0) = 3 0 + =. The range of the function f : R R, f () = 3 + is (, ). 0 = 3 = = 3 + Sketch the graphs of each of the following: a = 3 b = 3 c = 3 d = 3 + 4 a The graph of = 3 is obtained from the graph of = 3 badilation of factor from the -ais. The horizontal asmptote for both graphs has equation = 0. b The graph of = 3 is obtained from the graph of = 3 badilation of factor from the -ais. (In the notation introduced in Chapter 6, write it as (, ) (, ). Then describe the transformation as = and =, and hence = and =. The graph of = 3 is mapped to the graph of = 3 ) The horizontal asmptote for both graphs has equation = 0. = 0 = 0 0 0, 3 = 3 (, 6) (, 3) = 3 (, 3) = 3

Chapter 5 Eponential Functions and Logarithms 43 Eample 4 Eample 5 Eample 6 c The graph of = 3 is obtained from the graph of = 3 b a dilation of factor from the -ais. (In the notation introduced in Chapter 6, write it as (, ) (, ).) Then describe the transformation as = and =, and hence = and =. The graph of = 3 is mapped to the graph of = 3 ). d The graph of = 3 + 4is obtained from the graph of = 3 b a dilation of factor from the -ais, followed b a reflection in the -ais then b a translation of 4 units in the positive direction of the -ais. Eercise 5A = 0 0 0 = 3 (, 3) 3 (, 3) = 3 = 3 = 4 = 3 + 4 Using a calculator, plot the graphs of the following and comment on the similarities and differences between them: a =.8 b =.4 c = 0.9 d = 0.5 Using a calculator, plot the graphs of the following and comment on the similarities and differences between them: a = 3 b = 5 3 c = 3 d = 5 3 3 Using a calculator plot the graph of = for [ 4, 4] and hence find the solution of the equation = 4. 4 Using a calculator plot the graph of = 0 for [ 0.4, 0.8] and hence find the solution of the equation 0 = 6. 5 Sketch the graphs of each of the following functions. Give equations of asmptotes and -ais intercepts, and state the range of each of the functions. (-ais intercepts need not be given.) a f : R R, f () = 3 + b f : R R, f () = 3 3 c f : R R, f () = 3 d f : R R, f () = 3 + ( ) e f : R R, f () = + f f : R R, f () = 3 6 Sketch the graphs of each of the following: a = 5 b = 3 3 c = 5 d = 3 +

44 Essential Mathematical Methods &CAS 5. Reviewing rules for eponents (indices) We shall review the rules for manipulating the eponential epression a,where a is a real number called the base, and is a real number called the eponent or inde. Other words which are snonms for inde are power and logarithm. Multiplication: a m a n If m and n are positive integers then a m = a a a a m terms and a n = a a a n terms a m a n = (a a a a) (a a a) m terms n terms = (a a a a) m + n terms = a m + n Rule To multipl two numbers in eponent form with the same base, add the eponents. Eample 7 a m a n = a m + n Simplif each of the following: a 3 b 3 4 c + d 3a b 3 4a 3 b a 3 = 3 + b 3 4 = 4 3 = 5 = 6 4 c + = + + d 3a b 3 4a 3 b = 3 4a a 3 b 3 b = + = a 5 b 5 Using the TI-Nspire b, c and d can be simplified as shown.

Chapter 5 Eponential Functions and Logarithms 45 Using the Casio ClassPad c and d Enter ( + ) and 3a b 3 4a 3 b 3 in the entr line. Division: a m a n If m and n are positive integers and m > n m terms then a m a n = a a a...a a a a...a n terms = a a a...a (b cancelling) (m n) terms = a m n Rule To divide two numbers in eponent form with the same base, subtract the eponents. a m a n = a m n Define a 0 = for a 0 and a n = for a 0. an Note: Rule and Rule also hold for negative indices m, n for a 0. Foreample: 4 = 4 = = (i.e. 4 + ( ) ) 4 = 4 = 4 = 6 (i.e. 4 ) 3 3 = 3 3 = = 0 (i.e. 3 3 )

46 Essential Mathematical Methods &CAS Eample 8 Simplif each of the following: a 4 3 b b4 b + 6a 5 b 4a 4 b 3 c b 8ab a 4 3 = 4 3 b b4 b + = b b = = b 3 + c 6a5 b 4a 4 b 3 = 6 4 a 5 + 4 b + 3 8ab 8 = 8a 8 b 3 Raising the power: (a m ) n Consider the following: 4 + + ( 3 ) = 3 3 = 3 + 3 = 6 = 3 (4 3 ) 4 = 4 3 4 3 4 3 4 3 = 4 3 + 3 + 3 + 3 = 4 = 4 3 4 (a ) 5 = a a a a a = a + + + + = a 0 = a 5 In general (a m ) n = a m n. Rule 3 To raise the power of a to another power, multipl the indices. (a m ) n = a m n This rule holds for all integers m and n. Products and quotients ( a ) n. Consider (ab) n and b (ab) n = (ab) (ab) (ab) n terms = (a a a) (b b b) n terms n terms = a n b n Rule 4 (ab) n = a n b n ( a ) n a = b b a b a b = an b n

Rule 5 Eample 9 ( a b ) n = a n b n Chapter 5 Eponential Functions and Logarithms 47 Simplif the following, epressing the answers in positive eponent form: ( ) 4 a 8 3 3 6 4 3 b c d 3n 6 n 9 4 8 n 3 n a 8 = 8 = ( 3 ) c d = 6 b ( ) 4 = = 4 4 3 3 6 4 3 = 3 3 4 3 4 3 3 3 3 9 4 3 4 3 4 = 3 3 8 = 3 6 3n 6 n 8 n 3 = (3n 3 n ) (3 n n ) = 3n 3 n n 3n 3 n n ( ) 3 n = Eercise 5B Use the stated rule for each of the following to give an equivalent epression in simplest form: a 3 b 3 4 4 Rule 5 c 3 d 4 6 Rule 3 e (a 3 ) f ( 3 ) Rule 3 g () h ( 3 ) Rule 4 (also use Rule 3 for h) ( ) 3 ( 3 ) i j Rule 5 (also use Rule 3 for j)

48 Essential Mathematical Methods &CAS Simplif the following: Eample 7 a 3 4 b 4 4 3 8 c 3 4 9 7 3 d (q p) 3 (qp 3 ) e a b 3 (a 3 b ) 3 f ( 3 ) (4 4 ) 3 g m 3 p (m n 3 ) 4 (p ) h 3 a 3 b (a b ) 3 Simplif the following: Eample 8 a 3 5 b 6a5 b 4a 4 b 3 8ab ( ) ( ) 3 c 8() 3 d ( 3 3 ) 4 4 3 () 3 () 3 Eample 9 4 Simplif each of the following, epressing our answer in positive eponent form: a m 3 n p (mn p) 3 b 3 z ( 3 z) a b (ab ) 3 c z (a b ) a b 3 c 4 a n b 3 c n d e a b c 3 a n 3 b n c n 5 Simplif each of the following: a 3 4n 9 n 7 3n b n 8 n + 3 n 9 n 3 c 3 n 6 3 n + d n 9 n 5 n 5 n 6 3 4 e f 6 n 5 n + 3 + 6 n 9 3 g 7 n 8 n 6 n h 3n 9 n + 8 5 3 7 i 7 n 9 7 8 6 Simplif and evaluate: (8 3 ) 4 a b (5)3 (8) 4 7 3 c ( ) (5) 9 5.3 Rational eponents a n, where n is a natural number, is defined to be the n th root of a, and denoted n a. If a 0 then a n is defined for all n N.Ifa < 0 then n a is onl defined for n odd. (Remember that onl real numbers are being considered.) a n = n ( ) a, with a n n = a Using this notation for square roots: a = a = a Further, the epression a can be defined for rational eponents, i.e. when = m n,where m and n are integers, b defining ( ) a m n = a m n

Chapter 5 Eponential Functions and Logarithms 49 Eample 0 Evaluate: a 9 b 6 5 c 64 3 a 9 = = = ( ) b 6 5 = 6 5 ( ) 5 = 6 = 4 5 = 04 9 3 9 c 64 3 = 64 3 = ( 64 3 ) = ( 3 ) = 64 4 = 6 Earlier we stated the rules: a m a n = a m + n a m a n = a m n (a m ) n = a m n where m and n are integers. Note: These rules are applicable for all rational eponents. m n a q a p = a m Eample Simplif: n a q a p = a ( ) m n p m a q = a a 3 4 6 4 6 3 4 m q + n p m q n p q n p ( b ( ) a 3 4 6 4 = 3 4 3 4 ( 6 3 4 ) 6 3 4 3 = 3 4 3 4 = 3 4 3 4 3 3 = 3 3 4 4 3 4 = 3 3 4 9 4 ( ) b ( ) 4 = 3 4 = 3 5 3 ) 4

40 Essential Mathematical Methods &CAS Eample 0 Eample Eercise 5C Evaluate each of the following: a 5 3 b 43 3 5 c 8 d 64 3 ( ) 3 e f 3 5 g 5 3 h 3 4 5 8 ( i 000 4 3 j 0 000 3 4 k 8 3 7 4 l 5 Simplif: 3 a a b ( ab 3 b (a b) 3 d 3 4 4 6 3 4 e 3 Simplif each of the following: b 3 ) ( 3 ) ( 3 3 ) 3 3 3 4 f c 45 3 ) 3 9 3 4 5 3 ( ) 3 ( 5 a 3 a 5 a ( ) b ( ) c ( + ) + d ( ) 3 e + f (5 + ) 3 5 + 5.4 Solving eponential equations and inequations Method Epress both sides of the equation as eponents to the same base and then equate the eponents since, if a = a then =. Eample Find the value of for which: a 4 = 56 b 3 = 8 c 5 4 = 5 + a 4 = 56 4 = 4 4 = 4 Eample 3 b 3 = 8 3 = 3 4 = 4 = 5 ) 5 c 5 4 = 5 + = (5 ) + = 5 + 4 4= + 4 4 = 8 = Solve 9 = 3 7.

Chapter 5 Eponential Functions and Logarithms 4 (3 ) = 3 7 Let = 3 Then = 7 + 7 = 0 ( 3)( 9)= 0 Therefore 3= 0 or 9= 0 = 3 or = 9 Hence 3 = 3 or 3 = 3 and = or = Using the TI-Nspire to Eample 3 Use solve( ) from the Algebra menu (b 3). Eample 4 Solve 5 = 0 correct to decimal places. Press / to obtain the answer as a decimal number. =.43 (correct to decimal places)

4 Essential Mathematical Methods &CAS Eample Using the Casio ClassPad to Eample 3 Enter and highlight the equation in, then select Interactive Equation/inequalit solve and set the variable as. Solve 5 = 0 correct to decimal places. To answer the question as required, ou ma need to highlight the answer and click to convert from eact to decimal approimation. =.43 (correct to decimal places) of inequalities The propert that a > a >,where a (, ), and a > a < when a (0, ) is used. Eample 5 Solve for in each of the following: a 6 > b 3 + < 6 a 4 > 4 > b 3 + < 4 3 + < 4 > 3 < 5 4 > 5 3 Note: The CAS calculator can be used to help visualise the inequalit. Plot the graph of = 6 and find b using intersect to solve 6 >. Eercise 5D Solve for in each of the following: a 3 = 7 b 4 = 64 c 49 = 7 d 6 = 8 e 5 = 5 f 5 = 65 g 6 = 56 h 4 = i 5 = 64 5

Eample 3 Eample 4 Eample 5 Chapter 5 Eponential Functions and Logarithms 43 Solve for n in each of the following: a 5 n 5 n = 5 b 3 n 4 = c 3 n = 8 d 3n 9 = n e 33n 9 n + = 7 f 3n 4 n = 6 g n 6 = 8 n h 9 3n + 3 = 7 n i 4 n + = 8 n j 3 n + = 8 4n k 5 n + = 5 390 65 l 5 4 n = 5 6 n m 4 n = 048 3 Solve the following eponential equations: a 4 + = 3 b 3 9 = 43 c (7 3 ) = 7 3 4 Solve for : a 4( ) = 8( ) 4 b 8( ) 0( ) + = 0 c 3 8( ) + 4 = 0 d 9 4(3 ) + 3 = 0 5 Use a calculator to solve each of the following, correct to decimal places: a = 5 b 4 = 6 c 0 = 8 d 0 = 56 6 Solve for in each of the following: a 7 > 49 b 8 > c 5 5 d 3 + < 8 e 9 + < 43 f 4 + > 64 g 3 8 5.5 Logarithms Consider the statement 3 = 8. This ma be written in an alternative form: log 8 = 3 which is read as the logarithm of 8 to the base is equal to 3. In general, if a R + \ {} and R then the statements a = n and log a n = are equivalent. a = is equivalent to log a = Further eamples: 3 = 9isequivalent to log 3 9 = 0 4 = 0 000 is equivalent to log 0 0 000 = 4 a 0 = isequivalent to log a = 0 Eample 6 Without the aid of a calculator evaluate the following: a log 3 b log 3 8

44 Essential Mathematical Methods &CAS a Let log 3 = = 3 = 5 Therefore = 5, giving log 3 = 5. Laws of logarithms b Let log 3 8 = 3 = 8 3 = 3 4 Therefore = 4, giving log 3 8 = 4. The inde rules are used to establish other rules for computations with logarithms. Let a = m and a = n,where m, n and a are positive real numbers. mn = a a = a + log a mn = + and, since = log a m and = log a n,itfollows that: Foreample: log a mn = log a m + log a n Rule log 0 00 + log 0 5 = log 0 (00 5) = log 0 000 = 3 m n = a a = a ( m ) log a = and so: n ( m ) log a = log n a m log a n Rule Foreample: ( 3 log 3 log 8 = log 8 = log 4 = ( ) 3 If m =, log a = log n a log a n = log a n Therefore ( ) log a = log n a n Rule 3 4 m p = (a ) p = a p log a (m p ) = p )

Chapter 5 Eponential Functions and Logarithms 45 and so log a (m p ) = p log a m Rule 4 Foreample: 3log 3 4 = log 3 (4 3 ) = log 3 64 Eample 7 Without using a calculator simplif the following: ( ) 6 log 0 3 + log 0 6 log 0 5 ( ) ( 6 6 log 0 3 + log 0 6 log 0 = log 5 0 3 + log 0 6 log 0 5 ( ) 36 = log 0 9 + log 0 6 log 0 5 ( = log 0 9 6 5 ) 36 = log 0 00 = Eample 8 Solve each of the following equations for : a log 5 = 3 b log 5 ( + ) = c log ( + ) log ( ) = 4 d log 3 ( ) + log 3 ( + ) = a log 5 = 3 = 5 3 = 5 b log 5 ( + ) = + = 5 + = 5 = 4 = )

46 Essential Mathematical Methods &CAS Eample 6 Eample 7 c log ( + ) log ( ) = 4 ( ) + Then log = 4 + = 4 + = 6( ) Then 7 = 4 7 Therefore 4 = d log 3 ( ) + log 3 ( + ) = Therefore log 3 [( )( + )] = which implies = 3 and =± But the epression is not defined for =, therefore =. Eercise 5E Use the stated rule for each of the following to give an equivalent epression in simplest form: a log 0 + log a b log 0 5 + log 0 Rule c log 9 log 4 d log 0 log 5 Rule ( ) ( ) e log 5 f log 6 5 5 Rule 3 g log (a 3 ) h log (8 3 ) Rule 4 Without using a calculator evaluate each of the following: ( ) a log 3 7 b log 5 65 c log ( ) 8 d log 4 e log 4 f log 0.5 64 g log 0 0 000 h log 0 0.000 00 i 3log 5 5 j 4log 6 k log 3 9 l 4log 6 4 3 Without using a calculator simplif each of the following: a log 0 6 + log 0 5 b log 6 + log 8 c log 8 + log 3 45 log 3 5 d log 4 3 log 9 7 e log b b 3 log b b f log a + log a 3 g log 8 + log 8 h 3 log a a log a a

Eample 8 Chapter 5 Eponential Functions and Logarithms 47 4 Solve for : a log 3 9 = b log 3 = 3 c log 5 = 3 d log 0 = log 0 4 + log 0 e log 0 + log 0 5 + log 0 log 0 3 = f log 0 = log 0 36 log 0 3 g log 64 = h log 5 ( 3) = 3 i log 3 ( + ) log 3 = j log 0.0 = 5 Solve each of the following for. ( ) a log = b log 4 ( ) = 3 5 c log 4 ( + ) log 4 6 = d log 4 (3 + 4) + log 4 6 = 5 e log 3 ( 3 ) = 0 f log 3 ( 3 + ) = 0 6 If log 0 = a and log 0 = c,epress log 0 00 3 in terms of a and c. ( ab ) ( c ) 7 Prove that log 0 + log c 0 log ab 0 (bc) = 0. ( ) ( ) ( ) 490 7 8 If log a + log 3 a log 97 a = log 9 a (k), find k. 9 Solve each of the following equations for : a log 0 ( + 8) = log 0 b log 0 (5) log 0 (3 ) = c 3log 0 ( ) = log 0 8 d log 0 (0) log 0 ( 8) = e log 0 5 + log 0 ( + ) = + log 0 ( + 7) f + log 0 ( + ) = log 0 ( + ) + log 0 (5 + 8) 5.6 Using logarithms to solve eponential equations and inequations In Section 5.4 two methods were shown for solving eponential equations. If a R + \ {} and R then the statements a = n and log a n = are equivalent. This defining propert of logarithms ma be used in the solution of eponential equations. Method 3 Using logarithms Eample 9 Solve for if =.

48 Essential Mathematical Methods &CAS Take either the log 0 or log e (since these are the onl logarithmic functions available on our calculator) of both sides of the equation: Then log 0 = log 0 i.e. log 0 = log 0 Therefore = log 0 log 0 Use our calculator to evaluate both the log 0 and log 0. Then =.04 0.30 (to 3 decimal places) 3.46 (to decimal places) Eample 0 Solve 3 = 8. log 0 3 = log 0 8 ( ) log 0 3 = log 0 8 = log 0 8 log 0 3 =.447 0.477 3.033 Eample Solve {: 0.7 0.3}. = 4.033 =.07 (to 3 decimal places) Taking log 0 of both sides: log 0 0.7 log 0 0.3 log 0 0.7 log 0 0.3 log 0 0.3 Note change of direction of inequalit as log log 0 0.7 0 0.7 < 0. 0.59 0.549 3.376

Chapter 5 Eponential Functions and Logarithms 49 Eample 0 Eample Eample Eample Sketch the graph of f () = 0 4, giving the equation of the asmptote and the aes intercepts. f (0) = 0 0 4 = 4 = As, 4from the positive side, or as, 4 +. The equation of the horizontal asmptote is = 4. For the -ais intercept consider 0 4 = 0. Hence 0 = 4 and thus 0 =. Therefore = log 0 = 0.300 (correct to 4 decimal places) Eercise 5F Solve each of the following equations correct to decimal places: 0 log 0 () a = 7 b = 0.4 c 3 = 4 d 4 = 3 e = 6 f 0.3 = g 5 = 3 h 8 = 005 + i 3 = 0 j 0. + = 0.6 Solve for.givevalues correct to decimal places if necessar: a > 8 b 3 < 5 c 0.3 > 4 d 3 7 e 0.4 0.3 = 4 3 For each of the following sketch the graph of = f (), giving the equation of the asmptote and the aes intercepts: a f () = 4 b f () = 3 6 c f () = 3 0 5 d f () = 0 + 4 e f () = 3 + 6 f f () = 5 6 4 In the initial period of its life a particular species of tree grows in the manner described b the rule d = d 0 0 mt,where d is the diameter of the tree, in cm, t ears after the beginning of this period. The diameter is 5 cm after ear and 80 cm after 3 ears. Calculate the values of the constants d 0 and m.

430 Essential Mathematical Methods &CAS 5.7 Graph of = log a, where a > A table of values for = log 0 is given below (the values are correct to decimal places). Use our calculator to check these values. 0.8 0.6 0.4 0. 0 0. 0.4 0.6 0.8.0 0. 3 4 5 = log 0 0 0.30 0.48 0.60 0.70 = log 0 3 4 5 Domain = R +, Range = R 0 = 0 = = log 0 The graph of = log 0 is the reflection in the line = of the graph of = 0. Note that log 0 = 0as0 0 = and as 0 +,. The inverse of a one-to-one function was introduced in Section 6.7. The function f : (0, ) R, f () = log 0 is the inverse function of f : R R, f () = 0 In general, = log a is the rule for the inverse of the function with rule = a,where a > 0 and > 0. Note that log a (a ) = for all and a log a = for positive values of, asthe are inverse functions. For a >, the graphs of logarithm functions all have this same shape. Properties of = a, a > domain = R range = R + a 0 =, 0 + Properties of = log a, a > domain = R + range = R log a = 0 0 +, 0 = a = = log a Eample 3 Find the inverse of: a = 0 b = log 0 ()

a = 0 Interchanging and gives = 0 Therefore = log 0 and Eample 4 = log 0 Chapter 5 Eponential Functions and Logarithms 43 b = log 0 () Interchanging and gives = log 0 () Therfore 0 = and = 0 Find the inverse of each of the following: a f () = + 3 b f () = log ( ) c f () = 5 + 3 a Let = + 3 b Let = log ( ) Interchanging and gives Interchanging and gives = + 3 = log ( ) 3 = = and = log ( 3) and = + f () = log ( 3) f () = + domain of f = (3, ) domain of f = R c Let = 5 + 3 Interchanging and gives = 5 + 3 3 = 5 ( ) 3 and = log 5 ( ) 3 f () = log 5 domain of f = (3, ) Transformations can be applied to the graphs of logarithm functions. This is shown in the following eample. Eample 5 Sketch the graphs of each of the following. Give the maimal domain, the equation of the asmptote and the aes intercepts. a f () = log ( 3) b f () = log ( + ) c f () = log (3)

43 Essential Mathematical Methods &CAS Eample 3 Eample 4 a f () = log ( 3) For the function to be defined 3 > 0, i.e. > 3 The maimal domain is (3, ) -ais intercept: log ( 3) = 0 implies 3 = 0, i.e. = 4 Asmptote: 3 +, b f () = log ( + ) For the function to be defined + > 0, i.e. > The maimal domain is (, ) f(0) = log () = -ais intercept: log ( + ) = 0 implies + = 0, i.e. = Asmptote: +, c f () = log 3 For the function to be defined 3 > 0, i.e. > 0 The maimal domain is (0, ) For the -ais intercept log (3) = 0 implies 3 = 0, i.e. = 3 Asmptote: 0 +, Eercise 5G 0 = 0 3 = 0 = 3 Sketch the graph of each of the following and state the domain and range for ( each: ) a = log 0 () b = log 0 c = log 0 d = log 0 (3) e = log 0 f = log 0 ( ) Determine the inverse of each of the following: a = 0 0.5 b = 3log 0 c = 0 3 d = log 0 (3) 3 Find the rule for the inverse function of each of the following: a f () = 3 + b f () = log ( 3) c f () = 4 3( + ) d f () = 5 e f () = log (3) f f () = log 3 g f () = log ( + 3) h f () = 5 3 0 4

Eample 5 Chapter 5 Eponential Functions and Logarithms 433 4 Sketch the graphs of each of the following. Give the maimal domain, the equation of the asmptote and the aes intercepts: a f () = log ( 4) b f () = log ( ( + 3) c f () = log () ) d f () = log ( + ) e f () = log f f () = log ( ) 3 5 Use a calculator to solve each of the following equations correct to decimal places: a = b log 0 () + = 0 6 Use a calculator to plot the graphs of = log 0 ( ) and = log 0 for [ 0, 0], 0. 7 On the same set of aes plot the graph of = log 0 ( ) and = log 0 for (0, 0]. 8 Use a calculator to plot the graphs of = log 0 () + log 0 (3) and = log 0 (6 ) 9 Find a and k such that the graph of = a0 k passes through the points (, 6) and (5, 0). 5.8 Eponential models and applications Fitting data Using a TI-Nspire calculator It is known that the points (, 6) and (5, 96) lie on a curve with equation = a b. Define f () = a b. Then use solve( f () = 6 and f (5) = 96, {a, b}) a > 0 and b > 0. The result is as shown. The equations can also be solved for a and b b hand : a b = 6 and a b 5 = 96 Divide the second equation b the first to obtain b 4 = 6. Hence b =. Substitute in the first equation to obtain a = 3. There are man practical situations in which the relationship between variables is eponential.

434 Essential Mathematical Methods &CAS Eample 6 Takearectangular piece of paper approimatel 30 cm 6 cm. Fold the paper in half, successivel, until ou have folded it five times. Tabulate the times folded, f, and the number of creases in the paper, C. Times folded, f 0 3 4 5 Creases, C 0 3 7 5 3 The rule connecting C and f is C = f, f N {0}. Eample 7 C 30 5 0 5 0 5 0 3 4 5 The table below shows the increase in weight of Somu, an orang-utan born at the Eastern Plains Zoo. Draw a graph to show Somu s weight increase for the first si months. Months, m 0 3 4 5 6 Weight, w kg.65.7. 3.0 3.7 4. 4.8 Plotting these values: w (kg) 4 3 Graph showing Somu s weight increase Graph of w =.65(.) m Note: It is appropriate in this case to form a continuous line. 0 3 4 5 6 m (months) f

Chapter 5 Eponential Functions and Logarithms 435 We shall now plot, on the same set of aes, the graph of the eponential function w =.65(.) m,0 m 6. Tableofvalues m 0 3 4 5 6 w.65.98.38.85 3.4 4. 4.93 It can be seen from the graphs that the eponential model w =.65(.) m approimates to the actual weight gain and would be a useful model to predict weight gains for an future orang-utan births at the zoo. This model describes a growth rate for the first 6 months of 0% per month. This problem can also be attempted with a CAS calculator. Using the TI-Nspire Enter the data in either a Calculator application as lists or in a Lists & Spreadsheet application as shown. Choose Eponential Regression (b 4A) from the list of available regressions and complete as shown. Use the tab ke (e)tomove between the cells, use the selection tool ( )toopen a cell, and use the up/down arrows ( )to move to the entr to be selected. Select this entr using the selection tool ( ). This now gives the values of a and b, and the equation has been entered in f ().

436 Essential Mathematical Methods &CAS The curve can be shown in a Graphs & Geometr application together with the Scatter Plot (b 34) using an appropriate Window (b 4). Using the Casio ClassPad In enter the data in list and list as shown. In SetGraph Setting ensure that ou set the graph to Scatter and the lists to list and list. Now select Calc abeponential Reg and choose the settings shown. Eample 8 There are approimatel ten times as man red kangaroos as gre kangaroos in a certain area. If the population of gre kangaroos increases at a rate of % per annum while that of the red kangaroos decreases at 5% per annum, find how man ears must elapse before the proportions are reversed, assuming the same rates continue to appl.

Chapter 5 Eponential Functions and Logarithms 437 Let P = population of gre kangaroos at the start. number of gre kangaroos after n ears = P(.) n, and number of red kangaroos after n ears = 0P(0.95) n. When the proportions are reversed P(.) n = 0 [0P(0.95) n ] (.) n = 00(0.95) n Taking log 0 of both sides log 0 (.) n = log 0 00(0.95) n n log 0. = log 0 00 + n log 0 0.95 n 0.0453 = + n( 0.03) n = 0.0676 = 9.6 i.e. the proportions of kangaroo populations will be reversed b the 30th ear. Eponential growth The above two eamples are eamples of eponential change. In the following, A is a variable that is subject to eponential change. Let A be the quantit at time t. Then A = A 0 a,where A 0 is a positive constant and a is a real number. If a > the model represents growth. Ifa < the model represents deca. Phsical situations in which this is applicable include: the growth of cells population growth continuousl compounded interest radioactive deca cooling of materials. Consider a sum of mone, $0 000, invested at a rate of 5% per annum but compounded continuall. That is it is compounded at ever instant. If there are n compound periods in a ear, the interest paid per period is 5 %. Therefore n at the end of the ear the amount of the investment, A, is ( A = 0 000 + 5 ) n ( = 0 000 + ) n 00n 0n Enter the function Y = ( + /(0X))ˆX. Look at the table of values with an increment of one. It is found that the value approaches.057 (correct to 5 decimal places) for n large. Hence, for continuous compounding it can be written that A = 0 000 (.056...),where is the number of ears of the investment.

438 Essential Mathematical Methods &CAS Eample 7 Eercise 5H Find an eponential model of the form = ab to fit the following data. 0 4 5 0.5 0.5 0.7 0.09 0.006 Find an eponential model of the form p = ab t to fit the following data t 0 4 6 8 p.5 4.56 8.3 5. 7.56 3 A sheet of paper 0. mm thick is cut in half and one piece is stacked on top of the other. a If this process is repeated complete the following table: Cuts, n Sheets Total thickness, T (mm) 0 0. 0.4 4 0.8 3 8. 0. b Write down a formula which shows the relationship between T and n. c Draw a graph of T against n for n 0. d What would be the total thickness, T, after 30 cuts?.. 4 The populations (in millions), p and q, oftwo neighbouring American states, P and Q, over a period of 50 ears from 950 are modelled b functions p =. 0.08t and q =.7 0.04t,where t is the number of ears since 950. a Plot the graphs of both functions using a calculator. b Find when the population of state P is: i equal to the population of state Q ii twice the population of state Q.

Chapter 5 Eponential Functions and Logarithms 439 Chapter summar Groups of eponential functions Eamples: = a, a > = a, 0 < a < To multipl two numbers in eponent form with the same base, add the eponents: a m a n = a m + n To divide two numbers in eponent form with the same base, subtract the eponents: a m a n = a m n To raise the power of a to another power, multipl the eponents: (a m ) n = a m n If a = a then =. For rational eponents: a n = n a ( ) and a m n = a m n For a R + \ {} and R: a = is equivalent to log a = Laws of logarithms ( m ) log a (mn) = log a m + log a n log a = log n a m log a n ( ) 3 log a = log n a n 4 log a (m p ) = p log a m Eponential equations can be solved b taking logarithms of both sides. e.g. If = then log 0 = log 0 and hence can be solved. Multiple-choice questions 8 3 4 3 = A B 0 C 6 D E 9 a b The epression (ab ) ab 3 6a = 0 A B a C a b 6 D E a b 6 b 6 ab 6 8ab 5 3 The function f : R R, f () = 3 has range A R B R \ { } C (, ) D (, ) E [, ) Review

440 Essential Mathematical Methods &CAS Review 4 The function f : R + R,where f () = log 3, has an inverse function f. The rule for f is given b A f () = B f () = 3 C f () = 3 D f () = 3 E f () = log ( 3 ) 5 If log 0 ( ) 3log 0 = log 0, then is equal to A 80 3 B + 8 3 C 60 D + 6 E 8 3 6 The solution of the equation 5 5 = 0 is equals A B C 5 5 log 0 D log 5 E 5 5 7 The equation of the asmptote of = 3log (5) + is A = 0 B = C = 3 D = 5 E = 8 Which of the following graphs could be the graph of the function f () = a + b, where a and b are positive? A B C D O O E O O 9 Which one of the following functions has a graph with a vertical asmptote with equation = b? A = log ( b) B = C = + b + b b D = + b E = b mh 0 The epression (3mh ) mh is equal to 3 8m 6 A mh 6 B 6m C 6m h 6 6 D E h 6 m h 6 8mh 5 O

Chapter 5 Eponential Functions and Logarithms 44 Short-answer questions (technolog-free) Simplif each of the following, epressing our answer with positive inde: a 6 b 8 m 3 n 4 a 3 b a b a c b 0 d m 5 n 6 (ab ) 4 e 6a 8 0a 7 8(a 3 ) m n f 4a g h 6a 9 (a) 3 (mn ) 3 i (p q ) j (a 4 ) 3 k 6a a 4 + a 8 l 5a 3a a Use logarithms to solve each of the following equations: a = 7 b = 7 c 0 = d 0 = 3.6 e 0 = 0 f 0 = 00 g 5 = 00 h = 0. 3 Evaluate each of the following: a log 64 b log 0 0 7 c log a a d log 4 ( ) e log 3 7 f log g log 0 0.00 h log 6 4 4 Epress each of the following as single logarithms: a log 0 + log 0 3 b log 0 4 + log 0 3 log 0 6 c log 0 a log 0 b d log 0 a 3 log 0 5 e log 0 + log 0 log 0 f log 0 a + 3log 0 b log 0 c 5 Solve each of the following for : a 3 (3 7) = 0 b ( 8)( )= 0 c + = 0 d + 3 = 0 6 Sketch the graph of: a =. b = 3. c = 5. d = + e = f = + 7 Solve the equation log 0 + log 0 log 0 ( + ) = 0 8 Given 3 = 4 = z, show that z = +. 9 Evaluate log + 3log 5 log 5 log 50. 0 a Given that log p 7 + log p k = 0, find k. b Given that 4 log q 3 + log q log q 44 =, find q. Solve: a 4 a+ = 6 a (for a) b log = 4 + log ( + 5) (for ) Review

44 Essential Mathematical Methods &CAS Review Etended-response questions This problem is based on the so-called Tower of Hanoi puzzle. Given a number of different sized discs, the problem is to move a pile of discs (where n, the number of discs, is ) to a second location (if starting at A then either to B or C) according to the following rules: Onl one disc can be moved at a time. A total of onl three locations can be used to rest discs. location A A larger sized disc cannot be placed on top of a smaller disc. The task must be completed in the smallest possible number of moves. location B location C a Using two coins complete the puzzle. Repeat first with three coins and then four coins and thus complete the table. Number of discs, n 3 4 Minimum number of moves, M b Work out the formula which shows the relationship between M and n. Use our formula to etend the table of values for n = 5, 6 and 7. c Plot the graph of M against n. d Investigate, for both n = 3 and 4, to find whether there is a pattern for the number of times each particular disc is moved. To control an advanced electronic machine, 87 different switch positions are required. There are two kinds of switches available: Switch : These can be set in 9 different positions. Switch : These can be set to 3 different positions. If n of switch tpe and n + ofswitch tpe are used, calculate the value of n to give the required number of switch positions. 3 Research is being carried out to investigate the durabilit of paints of different thicknesses. The automatic machine shown in the diagram is proposed for producing a coat of paint of a particular thickness. Thick laer Blade set to reduce thickness Thin laer The paint is spread over a plate and a blade sweeps over the plate reducing the thickness of the paint. The process involves the blade moving at three different speeds.

Chapter 5 Eponential Functions and Logarithms 443 a Operating at the initial setting the blade reduces the paint thickness to one-eighth of the original thickness. This happens n times. What fraction of the paint thickness remains? Epress this as a power of. b The blade is then reset so that it removes three-quarters of the remaining paint. This happens (n )times. At the end of this second stage epress the remaining thickness as a power of. c The third phase of the process involves the blade being reset to remove half of the remaining paint. This happens (n 3)times. At what value of n would the machine have to be set to reduce a film of paint 89 units thick to unit thick? 4 A hermit has little opportunit to replenish supplies of tea and so, to eke out supplies for as long as possible, he dries out the tea leaves after use and then stores the dried tea in an airtight bo. He estimates that after each re-use of the leaves the amount of tannin in the used tea will be half the previous amount. He also estimates that the amount of caffeine in the used tea will be one-quarter of the previous amount. The information on the label of the tea packet states that the tea contains 79 mg of caffeine and 8 mg of tannin. a Write down epressions for the amount of caffeine when the tea leaves are re-used for the first, second, third and nth times. b Do the same for the amount of tannin remaining. c Find the number of times he can re-use the tea leaves if a tea containing more than three times as much tannin as caffeine is undrinkable. 5 Anew tpe of red snthetic carpet was produced in two batches. The first batch had a brightness of 5 units and the second batch 0 units. After a period of time it was discovered that the first batch was losing its brightness at the rate of 5% per ear while the second lost its brightness at the rate of 6% per ear. a Write down epressions for the brightness of each batch after n ears. b A person bought some carpet from the first batch when it was a ear old and some new carpet from the second batch. How long would it be before the brightness of the two carpets was the same? 6 The value of shares in Compan X increased linearl over a two-ear period according to the model = 0.8 + 0.7t,where t is the number of months from the beginning of Januar 997 and $ is the value of the shares at time t. The value of shares in Compan Y increased over the same period of time according to the model = 0 0.03t,where $ is the value of these shares at time t months. The value of shares in Compan Z increased over the same period according to the model z =.7 log 0 (5( + )), where $z is the value of the shares at time t months. Use a calculator to sketch the graphs of each of the functions on the one screen. a Find the values of the shares in each of the three companies at the end of June 997. b Find the values of the shares in the three companies at the end of September 998. Review

444 Essential Mathematical Methods &CAS Review c d During which months were shares in Compan X more valuable than shares in Compan Y? Forhow long and during which months were the shares in Compan X the most valuable? 7 In 000 in a game park in Africa it was estimated that there were approimatel 700 wildebeest and that their population was increasing at 3% per ear. At the same time, in the park there were approimatel 850 zebras and their population was decreasing at the rate of 4% per ear. Use a calculator to plot the graphs of both functions. a b After how man ears was the number of wildebeest greater than the number of zebras? It is also estimated that there were 000 antelope and their numbers were increasing b 50 per ear. After how man ears was the number of antelope greater than the number of zebras? 8 Students conducting a science eperiment on cooling rates measure the temperature of a beaker of liquid over a period of time. The following measurements were taken. a b Time (min) 3 6 9 5 8 Temperature ( C) 7.5 59 49 45.5 34 8 3.5 Find an eponential model to fit the data collected. Use this model to estimate: i the initial temperature of the liquid ii the temperature of the liquid after 5 minutes. It is suspected that one of the temperature readings was incorrect. c Re-calculate the model to fit the data, omitting the incorrect reading. d Use the new model to estimate: i the initial temperature of the liquid ii the temperature of the liquid at t =. e If the room temperature is 5 C, find the approimate time at which the cooling of the liquid ceased. 9 The curve with equation = ab passes through the points (, ) and (, 5) a Find the values of a and b. b Let b = 0 z. i Take logarithms of both sides (base 0) to find z as an epression in. ii Find the value of k and a such that = a0 k passes through the points (, ) and (, 5). 0 a Find an eponential model of the form = a.b to fit the following data. 0 4 5 0 5 3 0 00 b Epress the model ou have found in a in the form = a0 k. c Hence find an epression for in terms of.