25 = 2. Remember to put the base lower than the g of log so you don t get confused 2. Write the log equation in exponential form 2.

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1 LOGARITHMS The logarithm of a number to a given base is the exponent to which that base must be raised in order to produce the number. For example: What is the exponent that must be raised to in order to produce? The answer is of course! Since =. In log form this would appear as log = 1 LESSON N = a k is called EXPONENTIAL FORM log a N = k is called LOG FORM where N > 0; a > 0 and a 1 s 1. Write the exponential expression in logarithmic form = 64 base exponent = number log 4 64 = 3 1. a b = c log a c = b Remember to put the base lower than the g of log so you don t get confused. Write the log equation in exponential form.1 log 1_ 1 = -3 log number = exponent base -3 = 1_ 1. log x 9 = x = 9 3. Solve for x: 3.1 log 16 4 = x 16 x = 4 4 x = 4 x = 1 x = 1 _ 3. log x 4 = x = 4 x = The Log Laws Change to exponential form Solve as a normal exponential equation Normally this would be x = ± but because of restrictions, x = In order to make life easier when working with logs there are several laws that can be used to simplify expressions. 009

2 Law 1: log k k = 1 k > 0, k 1 E.g. log 3 3 = 1; log 7 7 = 1 Law : log k 1 = 0 k > 0, k 1 E.g. lo g _ 7 1 = 0; log 1_ 1 = 0 Law 3: log k x m = m log k x k > 0, k 1, x > 0 E.g. log 3 3 = ; log 3 3 = Law 4: log k (xy) = log k x + log k y k > 0, k 1, x > 0, y > 0 E.g. log 3 (x)= log 3 () + log 3 (x) Law : log k (_ x y ) = log x log y k > 0, k 1, x > 0, y > 0 k k log 3 ( _ x ) = log 3 () log 3 (x) s 1. Use the log laws to expand as far as possible: log _ y 4 1a = log (y 4 ) log (1a) First undo the division = (log y 4 ) (log 1 + log a) Then undo the multiplication. Remember to use brackets = log y 4 log 3 log a Distribute the negative through and write numbers as exponents = 4 log y 3log log a Now deal with the powers = 4 log y 3(1) log a Remember log a a = 1 = 4 log y 3 log a. Express the following as the log of a single term: log log 1 + 3log + log 3 log = log log 1 + log 3 + log 3 log First use nlog a b = log a b n = log log 1 + log 8 + log 3 log = (log + log 8 + log 3) (log 1 + log ) = log ( 8 3) log ( 1) Now, organise all + and signs together = log (10) log (60) = log (_ ) = log = 1 3. Without the use of a calculator, evaluate where possible. log 3 81 _ 81 and 43 are both powers of the base 3 log 3 43 = _ log 3 34 log 3 3 Use nlog b = log a a bn, or log a a b = b = _ 4 log 3 3 log 3 3 Page

3 = _ 4 log log 7 4 log 4 log 7 = log 4 log 7 Simplify the logs using the log laws log 4 log 7 = log 4 log 7 log 4 log 7 = (log 4 log ) 7 Factorise and simplify (log 4 log 7 ) =. If log = p and log3 = q, express the following in terms of p and q..1 log(6) = log (3 ) = log(3) + log() = q + p. log ( _ 3) = log() log(3) = p - q.3 log (0,9) = log (_ 10) 9 = log 9 log 10 = log3 1 = log3 1 = q 1 Log equations: In this section we are going to cover: log10 = log = 1from log k K = 1 Note: When the base is not given like log10, the base is automatically 10. So log = log 10 and log x = log 10 x Solving logarithmic and exponential equations Problems in every day life Log equations can be solved through two different methods producing the same result. Method 1: Since the log of a number to any given base is unique, it follows that if log k A = log k B A = B E.g. log (x + ) = log 7 x + = 7 x = Method : We use the relationship between logs and exponents here log k A = B k B = A E.g. log (_ 1 16 ) = x x = _ Lesson 1 Algebra Page 3

4 x = -4 x = 4 1 Solve for x: Introduce log on right hand side using the rule: log a a = 1 and reduce to one log using the product law log (x ) + log (x 3) = 1 Use log laws to condense on LHS log (x )(x 3) = log (x )(x 3) = x x + 6 = x x + 4 = 0 (x 4)(x 1) = 0 x = 4 & x= 1 N/A Note : x = 1 is not a solution since (x ) in log (x ) and (x 3) in log (x 3) are negative when x = 1. Always check your solution to ensure that your answer for x does not cause the number to be a negative. log (x )(x 3) = 1 Reduce to one log using the product law 1 = (x )(x 3) Use method and convert to the exponential form as the equation has been reduced to the desired format = x x + 6 We now solve for x 0 = x x = (x 4)(x 1) x = 4 OR x = 1 N/A Real Life applications: There are many real life situations that involve logs and log calculations. The following examples give some real life applications of logs. s: 1. The energy of a radio active molecule is described by the equation below: E = 6.1(.71) t t ³ 0 Where t is the time in seconds from the start of its existence. 1.1 What is its initial energy? Its initial energy is when it first begins its life and therefore the time is 0. Substituting this value into the equation gives E = 6.1(.71) -(0) Page 4

5 = 6.1(1) Anything to the power of 0 is 1 = How long has it existed if its energy is 1.? (rounded off to two decimal places) Substituting this value in for the energy we can calculate the time that has passed as follows 1. = 6.1(.71) -t _ = (.71)-t (.71) -t = t = log t = t = 1.6 seconds. The measurement of noise is often described by its decibel value and is given by the equation below, where N0 represents normal noise and NI represents the noise we are trying to measure. Noise (in decibels) = 0log (_ N0) NI.1 How much noise is there in decibels if we measure 4 times more noise than usual? (Rounded off to two decimal places) This means that NO = 4NI and therefore _ NO NI = _ 4NI NI = 4 Thus substituting the values into the equation gives Noise (in decibels) = 0log(4) = 1.04decibels Using a calculator, rounding to d.p.. If the noise measured gives 100-decibels how much larger is the measured noise than normal noise? As the question asks how much larger the measured noise is to regular noise we can say that for the time being that it is x times larger where x is the value we are looking for. Therefore the noise itself is x times N0. The expression of _ NI is therefore = _ xn0 NO N0 Which actually gives: x We can then use the equation given, substituting the value of 100 in on the right for the value of decibels and the value of x for the argument of the log 100 = 0log(x) Dividing both sides by 0 = log(x) since: log(x) = log10(x) [base of 10] x = 10 Taking the log back to exponential form 009 Lesson 1 Algebra Page

6 x = Activity 3. A population of rabbits increases according to the following equation, where n represents the number of rabbits and t represents the time in years. n = 4(4. t ) 3.1 How many rabbits were there to start with? 3. How long will it take for the population to reach 4640? : 3.1 At the start t = 0 (time equals 0) n = 4(4, 0 ) n = 4 3. This means that n=4640. So substitute into the equation and solve for t Activity 4640 = 4(4. t ) _ = 4,t 4. t = 193, 3 t = log , 3 t = 3, years 1. Write in log form: = = 44,14. Write in exponential form:.1 log 1 = 3. log 7_ 3_ 8 = 3 3. Solve for x: 3.1 log x 6 = 4 Page 6

7 3. log 8 = x 4. Use the log laws to simplify: 4.1 log a 6a log a 3 log a 4. log _ log 10. If log 3 4 = k write in terms of k: log 3 16_ 3 6. Solve for x: 6.1 log (x 8) + log (x + ) = log 4 6. log 3 (x ) = 1 7. The intensity measure of a television is given by: I = 36log ( A _ Ao) Where A represents the measured amplitude of the television s intensity and Ao represents the normal amplitude of light intensity. 7.1 What is the light intensity if the television emits light at an intensity of 6. times that of normal light? 009 Lesson 1 Algebra Page 7

8 7. How much larger is the intensity of the television if it gives an intensity reading of 13,? s 1.1 log 3 81 = 4 1. log = = 1. (_ 3 _ 8 ) 3 = x 4 = x = _ x = 4 6 ( 3 ) x = 1 = 4 3x = 1 x = _ log a 6a log a 3 log a = Log _ 6a a (3)() = log a a = 1 4. log log 10 4 = log log 10 = log 10 = log. log 3 16 log 3 3 = log = log = k log (x 8)(x + ) = log 4 (x 8)(x + ) = 4 x 6x 16 = 4 x 6x 40 = 0 (x 10)(x + 4) = 0 Page 8

9 ` x = 10 AND x = 4 N/A since x = 4 causes the numbers of the logs to be negative 6. log 3 (x ) = = x = x 7.1 This means that the measured light intensity A= 6,Ao Substitute into the equation and solve for I I = 36log (_ 6,Ao Ao ) I = 36log(6.) I = 9,3 (1d.p.) 7. Substituting the value of 13. for I and working out the ratio of _ A Ao 13. = 36log (_ Ao) A _ = log (_ Ao) A log (_ Ao) A = 0,366 _ A Ao = _ A =,3 (1d.p.) Ao This means that the measured intensity is,3 times bigger than the normal amplitude. 009 Lesson 1 Algebra Page 9

Warm Up Lesson Presentation Lesson Quiz. Holt McDougal Algebra 2

4-5 Warm Up Lesson Presentation Lesson Quiz Algebra 2 Warm Up Solve. 1. log 16 x = 3 2 64 2. log x 1.331 = 3 1.1 3. log10,000 = x 4 Objectives Solve exponential and logarithmic equations and equalities.