5A Exponential functions

Transcription

1 Chapter 5 5 Eponential and logarithmic functions bjectives To graph eponential and logarithmic functions and transformations of these functions. To introduce Euler s number e. To revise the inde and logarithm laws. To solve eponential and logarithmic equations. To find rules for the graphs of eponential and logarithmic functions. To find inverses of eponential and logarithmic functions. To appl eponential functions in modelling growth and deca. ur work on functions is continued in this chapter. Man of the concepts introduced in Chapters and 3 domain, range, transformations and inverse functions are used in the contet of eponential and logarithmic functions. An eponential function has a rule of the form f ( = ka, where k is a non-zero constant and the base a is a positive real number other than. Eponential functions were introduced in Mathematical Methods Units & 2 and it was shown that there are practical situations where these functions can be applied, including radioactive deca and population growth. Some of these applications are further investigated in this chapter. We also introduce the eponential function f ( = e, which has man interesting properties. In particular, this function is its own derivative. That is, f ( = f (. Here we define the number e as n e = lim + n n We will show that limits such as this arise in the consideration of compound interest. ISBN Evans et al. 206

2 20 Chapter 5: Eponential and logarithmic functions 5A Eponential functions The function f ( = a, where a R + \{}, isaneponential function. The shape of the graph depends on whether a > or0< a <. Graph of f( =a for a > Graph of f( =a for 0 < a < (, a, a, a (0, (0, (, a Ke values are f ( =, f (0 = and f ( = a. a The maimal domain is R and the range is R +. The -ais is a horizontal asmptote. An eponential function with a > is strictl increasing, and an eponential function with 0 < a < is strictl decreasing. In both cases, the function is one-to-one. Graphing transformations of f( =a Translations If the translation (, ( + h, + k is applied to the graph of = a, then the image has equation = a h + k. The horizontal asmptote of the image has equation = k. The range of the image is (k,. Eample Sketch the graph and state the range of = Eplanation The graph of = 2 is translated unit in the positive direction of the -ais and 2 units in the positive direction of the -ais. The mapping is (, ( +, , 2 (, 3 (2, 4 Translation of ke points: (, 2 (0, (0, (, 3 (, 2 (2, 4 The range of the function is (2,.

3 5A Eponential functions 2 Reflections If a reflection in the -ais, given b the mapping (, (,, is applied to the graph of = a, then the image has equation = a. The horizontal asmptote of the image has equation = 0. The range of the image is (, 0. Eample 2 Sketch the graph of = 3. Eplanation The graph of = 3 is reflected in the -ais. -, - 3 (0, - (, -3 The mapping is (, (,. Reflection of ke points: (, 3 (, 3 (0, (0, (, 3 (, 3 If a reflection in the -ais, given b the mapping (, (,, is applied to the graph of = a, then the image has equation = a. This can also be written as = ( a a or =. The horizontal asmptote of the image has equation = 0. The range of the image is (0,. Eample 3 Sketch the graph of = 6. Eplanation The graph of = 6 is reflected in the -ais. The mapping is (, (,. (-, 6 Reflection of ke points: (, 6 (, 6 (0,, 6 (0, (0, (, 6 (, 6 Dilations For k > 0, if a dilation of factor k from the -ais, given b the mapping (, (, k, is applied to the graph of = a, then the image has equation = ka. The horizontal asmptote of the image has equation = 0. The range of the image is (0,.

4 22 Chapter 5: Eponential and logarithmic functions Eample 4 Sketch the graph of each of the following: a = 3 5 b = Eplanation a The graph of = 5 is dilated b factor 3 from the -ais. (, 5 The mapping is (, (,3. Dilation of ke points: 3, 5 (0, 3 (, 5 (, 3 5 (0, (0, 3 (, 5 (, 5 b The graph of = 8 is dilated b factor 5 from the -ais. 8, 5 The mapping is (, (, 5. Dilation of ke points:, 40 0, 5 (, 8 (, 40 (0, (0, 5 (, 8 (, 8 5 For k > 0, if a dilation of factor k from the -ais, given b the mapping (, (k,, is applied to the graph of = a, then the image has equation = a k. The horizontal asmptote of the image has equation = 0. The range of the image is (0,. Eample 5 Sketch the graph of each of the following: a = 9 2 a b = 3 2 Eplanation The graph of = 9 is dilated b factor 2 from the -ais. (2, 9 The mapping is (, (2,. Dilation of ke points: (, 9 ( 2, 9 2, (0, 9 (0, (0, (, 9 (2, 9

5 5A Eponential functions 23 b The graph of = 3 is dilated b factor 2 from the -ais., 3 2 The mapping is (, ( 2,. Dilation of ke points: (, 3 ( 2, 3, 2 3 (0, (0, (0, (, 3 ( 2,3 Note: Since 9 2 = ( 9 2 = 3, the graph of = 9 2 is the same as the graph of = 3. Similarl, the graph of = 3 2 is the same as the graph of = 9. A translation parallel to the -ais results in a dilation from the -ais. For eample, if the graph of = 5 is translated 3 units in the positive direction of the -ais, then the image is the graph of = 5 3, which can be written = Hence, a translation of 3 units in the positive direction of the -ais is equivalent to a dilation of factor 5 3 from the -ais. Combinations of transformations We have seen translations, reflections and dilations applied to eponential graphs. In the following eample we consider combinations of these transformations. Eample 6 Sketch the graph and state the range of each of the following: a = b = 4 3 c = 0 2 a (, 5 3 (0, 4 7, 2 Graph of = 2 + 3: The asmptote has equation = 3. The -ais intercept is = 4. The range of the function is (3,. Eplanation The graph of = is obtained from the graph of = 2 b a reflection in the -ais followed b a translation 3 units in the positive direction of the -ais. The mapping is (, (, + 3. For eample: (, 2 (, 7 2 (0, (0, 4 (, 2 (, 5

6 24 Chapter 5: Eponential and logarithmic functions b c, 3 3 (0, 0 3, 3 4 Graph of = 4 3 : The asmptote has equation =. The -ais intercept is 4 0 = 0. The range of the function is (,. -2 0, 2 0 (, -3 (2, -2 Graph of = 0 2: The asmptote has equation = 2. The -ais intercept is 0 2 = 2 0. The range of the function is (, 2. The graph of = 4 3 is obtained from the graph of = 4 b a dilation of factor 3 from the -ais followed b a translation unit in the negative direction of the -ais. The mapping is (, ( 3,. For eample: (, 4 ( 3, 3 4 (0, (0, 0 (, 4 ( 3,3 The graph of = 0 2is obtained from the graph of = 0 b a reflection in the -ais followed b a translation unit in the positive direction of the -ais and 2 units in the negative direction of the -ais. The mapping is (, (+, 2. For eample: (, 2 0 (0, 0 (0, (, 3 (, 0 (2, 2 Note: We can use the method for determining transformations for each of the graphs in Eample 6. Here we show the method for part c: - Write the equation as = Rearrange to 2 = 0. - We choose to write = 2 and =. - Hence = 2 and = +. Section summar Graphs of eponential functions: = a, a > = a, 0 < a <

7 5A 5A Eponential functions 25 For a R + \{}, the graph of = a has the following properties: The -ais is an asmptote. The -ais intercept is. The -values are alwas positive. There is no -ais intercept. Transformations can be applied to eponential functions. For eample, the graph of = a b( h + k, where b > 0 can be obtained from the graph of = a b a dilation of factor from the -ais b followed b the translation (, ( + h, + k. Eercise 5A Eample Eample 2, 3 Eample 4, 5 Eample For each of the following functions, sketch the graph (labelling the asmptote and state the range: a = b = 2 3 c = 2 +2 d = For each of the following, use the one set of aes to sketch the two graphs (labelling asmptotes: a = 2 and = 3 b = 2 and = 3 c = 5 and = 5 d =.5 and =.5 For each of the following functions, sketch the graph (labelling the asmptote and state the range: a = 3 2 b = 2 5 c = 2 3 d = 2 3 Sketch the graph and state the range of each of the following: a = b = c = For each of the following functions, sketch the graph (labelling the asmptote and state the range: a = 3 b = 3 + c = 3 d = ( 3 e = f = ( 3 6 For each of the following functions, sketch the graph (labelling the asmptote and state the range: a = ( 2 2 b = ( 2 c = ( For f ( = 2, sketch the graph of each of the following, labelling asmptotes where appropriate: a = f ( + b = f ( + c = f ( + 2 ( d = f ( e = f (3 f = f 2 g = 2 f ( + h = f ( 2

8 26 Chapter 5: Eponential and logarithmic functions 5A 8 For each of the following functions, sketch the graph (labelling the asmptote and state the range: a = 0 b = c = d = 0 e = f = A bank offers cash loans at 0.04% interest per da, compounded dail. A loan of $0 000 is taken and the interest paable at the end of das is given b C = [ (.0004 ]. a Plot the graph of C against. b Find the interest at the end of: i 00 das ii 300 das. c After how man das is the interest paable $000? d A loan compan offers $0 000 and charges a fee of $4.25 per da. The amount charged after das is given b C 2 = i Plot the graph of C 2 against (using the same window as in part a. ii Find the smallest value of for which C 2 < C. 0 If ou invest $00 at an interest rate of 2% per da, compounded dail, then after das the amount of mone ou have (in dollars is given b = 00(.02. For how man das would ou have to invest to double our mone? a i Graph = 2, = 3 and = 5 on the same set of aes. ii For what values of is 2 > 3 > 5? iii For what values of is 2 < 3 < 5? iv For what values of is 2 = 3 = 5? b Repeat part a for = ( 2, = ( 3 and = ( 5. c Use our answers to parts a and b to sketch the graph of = a for: i a > ii a = iii 0 < a < 5B The eponential function f( =e In the previous section, we eplored the famil of eponential functions f ( = a, where a R + \{}. ne particular member of this famil is of great importance in mathematics. This function has the rule f ( = e, where e is Euler s number, named after the eighteenth centur Swiss mathematician Leonhard Euler. Euler s number is defined as follows. Euler s number e = lim n ( + n n

9 5B The eponential function f( =e 27 To see what the value of e might be, we could tr large values of n and use a calculator to evaluate ( + n n, as shown in the table on the right. As n is taken larger and larger, it can be seen that ( + n n approaches a limiting value ( n ( + n n 00 (.0 00 = ( = ( = ( = ( = Like π, the number e is irrational: e = The function f ( = e is ver important in mathematics. In Chapter 9 ou will find that it has the remarkable propert that f ( = f (. That is, the derivative of e is e. Note: The function e can be found on our calculator. Graphing f( =e The graph of = e is as shown. (, 3 (, e = 3 = e = 2 The graphs of = 2 and = 3 are shown on the same set of aes., e, 2 (0, (, 2, 3 Eample 7 Sketch the graph of f : R R, f ( = e + 2. (, e 2 2 = e + 2 e 2 = 2 2 Eplanation To find the transformation: Write the image as + 2 = e +. We can choose = + 2 and = +. Hence = 2 and =. The mapping is (, (, 2 which is a translation of unit in the negative direction of the -ais and 2 units in the negative direction of the -ais. The asmptote has equation = 2. The -ais intercept is e 2.

10 28 Chapter 5: Eponential and logarithmic functions 5B Compound interest Assume that ou invest $P at an annual interest rate r. If the interest is compounded onl once per ear, then the balance of our investment after t ears is given b A = P( + r t. Now assume that the interest is compounded n times per ear. The interest rate in each period is r n. The balance at the end of one ear is P ( + r n n, and the balance at the end of t ears is given b A = P ( + r n We recognise that ( + n r n r = e lim n r nt (( r n rt = P + r n So, as n, we can write A = Pe rt. For eample, if $000 is invested for one ear at 5%, the resulting amount is $050. However, if the interest is compounded continuousl, then the amount is given b A = Pe rt = 000 e 0.05 = That is, the balance after one ear is $ Section summar Euler s number is the natural base for eponential functions: n e = lim ( + n n = Eercise 5B Skillsheet Sketch the graph of each of the following and state the range: Eample 7 a f ( = e + b f ( = e c f ( = e d f ( = e 2 e f ( = e 2 f f ( = 2e g h( = 2( + e h h( = 2( e i g( = 2e + j h( = 2e k f ( = 3e + 2 l h( = 2 3e 2 For each of the following, give a sequence of transformations that maps the graph of = e to the graph of = f (: a f ( = e +2 3 b f ( = 3e + 4 c f ( = 5e 2+ d f ( = 2 e e f ( = 3 2e +2 f f ( = 4e 2

11 5B 5B The eponential function f( =e 29 3 Find the rule of the image when the graph of f ( = e undergoes each of the following sequences of transformations: a a dilation of factor 2 from the -ais, followed b a reflection in the -ais, followed b a translation 3 units in the positive direction of the -ais and 4 units in the negative direction of the -ais b a dilation of factor 2 from the -ais, followed b a translation 3 units in the positive direction of the -ais and 4 units in the negative direction of the -ais, followed b a reflection in the -ais c a reflection in the -ais, followed b a dilation of factor 2 from the -ais, followed b a translation 3 units in the positive direction of the -ais and 4 units in the negative direction of the -ais d a reflection in the -ais, followed b a translation 3 units in the positive direction of the -ais and 4 units in the negative direction of the -ais, followed b a dilation of factor 2 from the -ais e a translation 3 units in the positive direction of the -ais and 4 units in the negative direction of the -ais, followed b a dilation of factor 2 from the -ais, followed b a reflection in the -ais f a translation 3 units in the positive direction of the -ais and 4 units in the negative direction of the -ais, followed b a reflection in the -ais, followed b a dilation of factor 2 from the -ais. 4 For each of the following, give a sequence of transformations that maps the graph of = f ( to the graph of = e : a f ( = e +2 3 b f ( = 3e + 4 c f ( = 5e 2+ d f ( = 2 e e f ( = 3 2e +2 f f ( = 4e 2 5 Solve each of the following equations using a calculator. Give answers correct to three decimal places. a e = + 2 b e = + 2 c 2 = e d 3 = e 6 a Using a calculator, plot the graph of = f ( where f ( = e. b Using the same screen, plot the graphs of: ( i = f ( 2 ii = f iii = f ( 3

12 220 Chapter 5: Eponential and logarithmic functions 5C Eponential equations ne method for solving eponential equations is to use the one-to-one propert of eponential functions: a = a implies =, for a R + \{} Eample 8 Find the value of for which: a 4 = 256 b 3 = 8 a 4 = = 4 4 b 3 = 8 3 = 3 4 = 4 = 4 = 5 When solving an eponential equation, ou ma also need to use the inde laws. Inde laws For all positive numbers a and b and all real numbers and : a a = a + ( a = b a b a a = a (a = a (ab = a b a = a a = a 0 = a Note: More generall, each inde law applies for real numbers a and b provided both sides of the equation are defined. For eample: a m a n = a m+n for a R and m, n Z. Eample 9 Find the value of for which = Eplanation = Epress both sides of the equation as powers with base 5. = ( = = Use the fact that 5 a = 5 b implies a = b. 4 = 8 = 2 To solve the equations in the net eample, we must recognise that the will become quadratic equations once we make a substitution.

13 5C 5C Eponential equations 22 Eample 0 Solve for : a 9 = a We can write the equation as (3 2 = Let = 3. The equation becomes 2 = = 0 ( 3( 9 = 0 = 3 or = 9 3 = 3 or 3 = 3 2 = or = 2 b b 3 2 = We can write the equation as (3 2 = Let = 3. The equation becomes 2 = = 0 ( 3( + 9 = 0 = 3 or = 9 3 = 3 or 3 = 9 The onl solution is =, since 3 > 0 for all. Section summar ne method for solving an eponential equation, without using a calculator, is first to epress both sides of the equation as powers with the same base and then to equate the indices (since a = a implies =, for an base a R + \{}. For eample: 2 + = = = 3 = 2 Equations such as = 0 can be solved b making a substitution. In this case, substitute = 3 to obtain a quadratic equation in. Eercise 5C Eample 8 Simplif the following epressions: a b c (3 4 d ( ( 2( 2 4 e (4 0 2 f 5( ( 3( ( g h ( i Solve for in each of the following: a 3 = 8 b 8 = 9 c 2 = 256 d 625 = 5 e 32 = 8 f 5 = 25 g 6 = 024 h 2 = i 5 =

14 222 Chapter 5: Eponential and logarithmic functions 5C Eample 9 3 Solve for n in each of the following: a 5 2n 25 2n = 625 b 4 2n 2 = c 4 2n = n 2 d = 27 e 92 n 22n 2 4 3n = 64 g 27 n 2 = 9 3n+2 h 8 6n+2 = 8 4n j 2 n 4 2n+ = 6 k (27 3 n n = 27 n 3 4 f i 2 n 4 = 8 4 n 25 4 n = 5 6 2n Eample 0 4 Solve for : a 3 2 2(3 3 = 0 b (5 50 = 0 c 5 2 0( = 0 d 2 2 = 6(2 8 e 8(3 6 = 2(3 2 f (2 = 64 g 4 2 5(4 = 4 h 3(3 2 = 28(3 9 i 7(7 2 = 8(7 5D Logarithms Consider the statement 2 3 = 8 This ma be written in an alternative form: log 2 8 = 3 which is read as the logarithm of 8 to the base 2 is equal to 3. For a R + \{}, the logarithm function with base a is defined as follows: a = is equivalent to log a = Note: Since a is positive, the epression log a is onl defined when is positive. Further eamples: 3 2 = 9 is equivalent to log 3 9 = = is equivalent to log = 4 a 0 = is equivalent to log a = 0 Eample Without the aid of a calculator, evaluate the following: a log 2 32 b log 3 8 a Let log 2 32 = b Let log 3 8 = Then 2 = 32 2 = 2 5 Therefore = 5, giving log 2 32 = 5. Then 3 = 8 3 = 3 4 Therefore = 4, giving log 3 8 = 4.

15 5D Logarithms 223 Note: To find log 2 32, we ask What power of 2 gives 32? To find log 3 8, we ask What power of 3 gives 8? Inverse functions For each base a R + \{}, the eponential function f : R R, f ( = a is one-to-one and so has an inverse function. Let a R + \{}. The inverse of the eponential function f : R R, f ( = a is the logarithmic function f : R + R, f ( = log a. log a (a = for all R a log a = for all R + Because the are inverse functions, the graphs of = a and = log a are reflections of each other in the line =. = log a = a = a = log a a > 0 < a < The natural logarithm Earlier in the chapter we defined the number e and the important function f ( = e. The inverse of this function is f ( = log e. Because the logarithm function with base e is known as the natural logarithm, the epression log e is also written as ln. The common logarithm The function f ( = log 0 has both historical and practical importance. When logarithms were used as a calculating device, it was often base 0 that was used. B simplifing calculations, logarithms contributed to the advancement of science, and especiall of astronom. In schools, books of tables of logarithms were provided for calculations and this was done up to the 970s. Base 0 logarithms are used for scales in science such as the Richter scale, decibels and ph. You can understand the practicalit of base 0 b observing: log 0 0 = log 0 00 = 2 log = 3 log = 4 log 0 0. = log = 2 log = 3 log = 4

16 224 Chapter 5: Eponential and logarithmic functions Laws of logarithms The inde laws are used to establish rules for computations with logarithms. Law : Logarithm of a product The logarithm of a product is the sum of their logarithms: log a (mn = log a m + log a n Proof Let log a m = and log a n =, where m and n are positive real numbers. Then a = m and a = n, and therefore For eample: mn = a a = a + Hence log a (mn = + = log a m + log a n. log log 0 5 = log 0 (200 5 = log = 3 (using the first inde law Law 2: Logarithm of a quotient The logarithm of a quotient is the difference of their logarithms: ( m log a = log n a m log a n Proof Let log a m = and log a n =, where m and n are positive real numbers. Then as before a = m and a = n, and therefore m n = a a = a (using the second inde law ( m Hence log a = = log n a m log a n. For eample: ( 32 log 2 32 log 2 8 = log 2 8 = log 2 4 = 2 Law 3: Logarithm of a power log a (m p = p log a m Proof Let log a m =. Then a = m, and therefore m p = (a p = a p (using the third inde law Hence log a (m p = p = p log a m. For eample: log 2 32 = log 2 (2 5 = 5

17 5D Logarithms 225 Law 4: Logarithm of m log a (m = log a m Proof Use logarithm law 3 with p =. For eample: ( log a 2 = loga (2 = log a 2 Law 5 log a = 0 and log a a = Proof Since a 0 =, we have log a = 0. Since a = a,wehavelog a a =. Eample 2 Epress the following as the logarithm of a single term: ( 6 2 log e 3 + log e 6 2 log e 5 ( 6 ( 6 2 log e 3 + log e 6 2 log e = log 5 e (3 2 + log e 6 log e 5 ( 36 = log e 9 + log e 6 log e 25 ( = log e = log e 00 2 Logarithmic equations Eample 3 Solve each of the following equations for : a log 2 = 5 b log 2 (2 = 4 c log e (3 + = 0 a log 2 = 5 b log 2 (2 = 4 c log e (3 + = 0 = 2 5 = 32 2 = = 7 = = e 0 3 = = 0

18 226 Chapter 5: Eponential and logarithmic functions Eample 4 Solve each of the following equations for : a log e ( + log e ( + 2 = log e (6 8 b log 2 log 2 (7 2 = log 2 6 a log e ( + log e ( + 2 = log e (6 8 log e ( ( ( + 2 = loge ( = = 0 ( 3( 2 = 0 = 3or = 2 Note: The solutions must satisf > 0, + 2 > 0 and 6 8 > 0. Therefore both of these solutions are allowable. b log 2 log 2 (7 2 = log 2 6 ( log 2 = log = 6 = = 42 = 42 3 Using the TI-Nspire Use solve from the Algebra menu as shown. Note that ln( = log e (. The logarithm with base e is available on the kepad b pressing ctrl e. Note: Logarithms with other bases are obtained b pressing the log ke ( ctrl 0 and completing the template. Using the Casio ClassPad For a logarithm with base e, usel from the Math keboard. Note that ln( = log e (. Enter and highlight the equation ln( + ln( + 2 = ln(6 8. Select Interactive > Equation/Inequalit > solve. Ensure the variable is set to. Note: For logarithms with other bases, tapvand complete the template. Eample 5 Solve each of the following equations for : a log e (2 + log e ( = 4 b log e ( + log e ( + =

19 5D 5D Logarithms 227 a log e (2 + log e ( = 4 b log e ( + log e ( + = ( 2 + ( log log e = 4 e ( ( + = log e ( 2 = 2 + = e4 2 = e Eample = e 4 ( (2 e 4 = (e 4 + = e4 + e 4 2 Solve the equation log 27 = 3 2 for. log 27 = 3 2 is equivalent to 3 2 = 27 ( 3 = 3 3 = 3 = 9 = ± e + But the original equation is not defined for = e + and so the onl solution is = e +. Section summar For a R + \{}, the logarithm function base a is defined as follows: a = is equivalent to log a = To evaluate log a ask the question: What power of a gives? For a R + \{}, the inverse of the eponential function f : R R, f ( = a is the logarithmic function f : R + R, f ( = log a. log a (a = for all a log a = for all positive values of Laws of logarithms ( m log a (mn = log a m + log a n 2 log a = log n a m log a n 3 log a (m p = p log a m 4 log a (m = log a m 5 log a = 0 and log a a = Eercise 5D Skillsheet Evaluate each of the following: ( Eample a log b log 2 c log ( d log 2 64 e log f log 2 28

20 228 Chapter 5: Eponential and logarithmic functions 5D Eample 2 2 Epress each of the following as the logarithm of a single term: a log e 2 + log e 3 b log e 32 log e 8 ( c log e 0 + log e 00 + log e 000 d log e + log 2 e 4 ( ( ( e log e + log 3 e + log 4 e f log 5 e (uv + log e (uv 2 + log e (uv 3 g 2 log e + 5 log e h log e ( + + log e ( log e ( 2 2 Eample 3 3 Solve each of the following equations for : a log 0 = 2 b 2 log 2 = 8 c log e ( 5 = 0 d log 2 = 6 e 2 log e ( + 5 = 6 f log e (2 = 0 g log e (2 + 3 = 0 Eample 4 4 Solve each of the following equations for : h log 0 = 3 i 2 log 2 ( 4 = 0 a log 0 = log log 0 5 b log e = log e 5 log e 3 c log e = 2 3 log e 8 d log e + log e (2 = 0 e 2 log e log e ( = log e ( Epress each of the following as the logarithm of a single term: a log log 0 3 b log 2 24 log 2 6 d + log 0 a 3 log 0 b e 2 log log log Without using our calculator, evaluate each of the following: c 2 log 0 a 2 log 0 b a log log 0 2 b log log 0 2 log 0 4 c log log2 + 2 log 2 2 d 2 log log e 4 log 0 2 log Simplif the following epressions: a log 3 ( 3 b log 2 2 log 2 + log 2 ( 2 c log e ( 2 2 log e ( log e ( + Eample 5 8 Solve each of the following equations for : a log e ( = 2 log e b log e (5 log e (3 2 = 9 Solve each of the following equations for : a log e + log e (3 + = b 8e e = 2 Eample 6 0 Solve each of the following equations for : ( a log 8 = 4 b log = 5 32 Solve 2 log e + log e 4 = log e ( Given that log a N = 2( loga 24 log a log a 3, find the value of N.

21 5E Graphing logarithmic functions 229 5E Graphing logarithmic functions The graphs of = e and its inverse function = log e are shown on the one set of aes., e = (0, = e (, e (, 0 e, (e, = log e The graphs of = log 2, = log e and = log 3 are shown on the one set of aes. (, 0 = log 2 = log e (2, = log 3 (3, (e, For each base a R + \{}, the graph of the logarithmic function f ( = log a has the following features: ( Ke values are f =, f ( = 0 and f (a =. a The maimal domain is R + and the range is R. The -ais is a vertical asmptote. A logarithmic function with a > is strictl increasing, and a logarithmic function with 0 < a < is strictl decreasing. In both cases, the function is one-to-one. Graphing transformations of f( = log a We now look at transformations applied to the graph of f ( = log a where a >. We make the following general observations: The graph of = log a (m n, where m > 0, has a vertical asmptote = n and implied ( n m domain m,. The -ais intercept is + n m. Eample 7 Sketch the graph of = 3 log e (2. This is obtained from the graph of = log e b a dilation of factor 3 from the -ais and a dilation of factor 2 from the -ais. The mapping is (, ( 2,3. (, 0 ( 2,0 (e, ( 2 e,3 e 2, 3 2, 0

22 230 Chapter 5: Eponential and logarithmic functions Eample 8 Sketch the graph and state the implied domain of each of the following: a = log 2 ( 5 + b = log 3 ( + 4 a The graph of = log 2 ( 5 + is obtained from the graph of = log 2 b a translation of 5 units in the positive direction of the -ais and unit in the positive direction of the -ais. The mapping is (, ( + 5, +. (, 0 (6, (2, (7, 2 The asmptote has equation = 5. When = 0, log 2 ( 5 + = 0 log 2 ( 5 = 5 = 2 5 (7, 2 (6, 5, 0 2 = 5 2 The domain of the function is (5,. b The graph of = log 3 ( + 4 is obtained from the graph of = log 3 b a reflection in the -ais and a translation of 4 units in the negative direction of the -ais. The mapping is (, ( 4,. (, 0 ( 3, 0 (3, (, The asmptote has equation = 4. When = 0, = log 3 ( = log 3 4 The domain of the function is ( 4,. (-, - -log 3 4 Eample 9 Sketch the graph of = 2 log e ( and state the implied domain. The graph of = 2 log e ( is obtained from the graph of = log e b a dilation of factor 2 from the -ais followed b a translation of 5 units in the negative direction of the -ais and 3 units in the negative direction of the -ais. 5 (0, 2 log e 5 3 ( 0.58, 0 The equation of the asmptote is = 5. The domain of the function is ( 5,. = 5

23 5E Graphing logarithmic functions 23 Ais intercepts When = 0, = 2 log e ( = 2 log e 5 3 When = 0, 2 log e ( = 0 log e ( + 5 = = e 3 2 = e Eponential and logarithmic graphs with different bases It is often useful to know how to go from one base to another. To change the base of log a from a to b (where a, b > 0 and a, b, we use the definition that = log a implies a =. Taking log b of both sides: log b (a = log b log b a = log b = log b log b a Since = log a, this gives: log a = log b log b a Hence the graph of = log a can be obtained from the graph of = log b b a dilation of factor log b a from the -ais. (, 0 (b, = log b = log a b, log b a Using properties of inverses, we can write a = b log b a. This gives: a = b (log b a (, b = b = a Hence the graph of = a can be obtained from the graph of = b b a dilation of factor log b a from the -ais. - log b a, b -, b (0,, b log b a

24 232 Chapter 5: Eponential and logarithmic functions 5E Eample 20 Find a transformation that takes the graph of = 2 to the graph of = e. We can write e = 2 log 2 e and so e = (2 log 2 e = 2 (log 2 e The graph of = e is the image of the graph of = 2 under a dilation of factor log 2 e from the -ais. Eercise 5E Eample 7 Eample 8 Eample Sketch the graph of each of the following: a = 2 log e (3 b = 4 log e (5 ( c = 2 log e (4 d = 3 log e 2 For each of the following functions, sketch the graph (labelling ais intercepts and asmptotes and state the maimal domain and range: a = 2 log e ( 3 b = log e ( c = 2 log e ( + d = 2 + log e (3 2 e = 2 log e ( + 2 f = 2 log e ( 2 g = log e ( + h = log e (2 i + = log e (4 3 Sketch the graph of each of the following. Label the ais intercepts and asmptotes. State the implied domain of each function. a = log 2 (2 b = log 0 ( 5 c = log 0 d = log 0 ( e = log 0 (5 f = 2 log 2 (2 + 2 g = 2 log 2 (3 h = log 0 ( i = 4 log 2 ( 3 j = 2 log 2 (2 6 k = log e (2 l = log e (3 2 4 Solve each of the following equations using a calculator. Give answers correct to three decimal places. a + 2 = log e b 3 log e (2 + = 2 +

25 5E 5F Determining rules for graphs of eponential and logarithmic functions a Using a calculator, plot the graph of = f ( where f ( = log e. b Using the same screen, plot the graphs of: ( i = f ( ii = f ( iii = f 3 iv = f (3 Eample 20 6 Find a transformation that takes the graph of = 3 to the graph of = e. 7 Find a transformation that takes the graph of = e to the graph of = 2. 5F Determining rules for graphs of eponential and logarithmic functions In previous chapters, we have determined the rules for graphs of various tpes of functions, including polnomial functions. In this chapter, we consider similar questions for eponential and logarithmic functions. Eample 2 The rule for the function with the graph shown is of the form = ae + b. Find the values of a and b. (0, 6 (3, 22 When = 0, = 6 and when = 3, = 22: 6 = ae 0 + b ( 22 = ae 3 + b (2 Subtract ( from (2: 6 = a(e 3 e 0 6 = a(e 3 a = 6 e 3 From equation (: b = 6 a = 6 6 e 3 = 6e3 22 e 3 ( The function has rule = 6 e 3 e + 6e3 22 e 3.

26 234 Chapter 5: Eponential and logarithmic functions Eample 22 The rule for the function with the graph shown is of the form = a log e ( + b. Find the values of a and b. (8, (5, 0 0 = a log e (5 + b ( = a log e (8 + b (2 From (: log e (5 + b = b = e 0 b = 4 Eplanation Form two equations in a and b b substituting into the rule = a log e ( + b: = 0 when = 5 = when = 8 Substitute in (2: = a log e 4 a = log e 4 The rule is = log e 4 log e ( 4. Eample 23 Given that = Ae bt with = 6 when t = and = 8 when t = 2, find A and b. 6 = Ae b ( 8 = Ae 2b (2 4 Divide (2 b (: 3 = eb 4 b = log e 3 Substitute in (: 6 = Ae log e = 4 3 A elog e A = 8 4 = 9 2 The rule is = 9 2 e ( loge 4 3 Eplanation Form two equations in A and b b substituting into the rule = Ae bt : = 6 when t = = 8 when t = 2 t. Note that = 9 2( e log e 4 3 = 4 3 since elog e a = a for all a > 0 4 t 3 = 9 ( 4 t 2 3

27 5F 5F Determining rules for graphs of eponential and logarithmic functions 235 Using the TI-Nspire Use menu > Algebra > Solve Sstem of Equations > Solve Sstem of Equations and complete as shown. Note: Do not use the e from the alpha kes; it will be treated as a variable. Using the Casio ClassPad Select the simultaneous equations template~. Enter the equations as shown: selectqfrom the Math keboard and select the parameters a, b from the Var keboard. Eercise 5F Eample 2 An eponential function has rule = a e + b and the points with coordinates (0, 5 and (4, are on the graph of the function. Find the values of a and b. Eample 22 2 A logarithmic function has rule = a log e ( + b and the points with coordinates (5, 0 and (0, 2 are on the graph of the function. Find the values of a and b. 3 The graph shown has rule = ae + b Find the values of a and b. (0, 6 = 4 4 The rule for the function for which the graph is shown is of the form = ae + b (, 4 Find the values of a and b. Eample 23 5 Find the values of a and b such that the graph of = ae b goes through the points (3, 50 and (6, 0.

28 236 Chapter 5: Eponential and logarithmic functions 5F 6 The rule for the function f is of the form f ( = ae + b Find the values of a and b. (0, 700 = Find the values of a and b such that the graph of = a log 2 + b goes through the points (8, 0 and (32, 4. 8 The rule of the graph shown is of the form = a log 2 ( b Find the values of a and b. = 5 (7, 3 9 Find the values of a and b such that the graph of = ae b goes through the points (3, 0 and (6, Find the values of a and b such that the graph of = a log 2 ( b passes through the points (5, 2 and (7, 4. The points (3, 0 and (5, 2 lie on the graph of = a log e ( b + c. The graph has a vertical asmptote with equation =. Find the values of a, b and c. 2 The graph of the function with rule f ( = a log e ( + b passes though the points ( 2, 6 and ( 4, 8. Find the values of a and b. 5G of eponential equations using logarithms Eample 24 If log 2 6 = k log 2 3 +, find the value of k. log 2 6 = k log = log 2 (3 k + log 2 2 = log 2 (2 3 k 6 = 2 3 k 3 = 3 k k =

29 5G of eponential equations using logarithms 237 Eample 25 Solve for if 2 =, epressing the answer to two decimal places. 2 = = log 2 = Therefore 3.46 correct to two decimal places. Eample 26 Solve 3 2 = 28, epressing the answer to three decimal places. 3 2 = 28 2 = log 3 28 Thus 2 = log = log = ( log correct to three decimal places Using the TI-Nspire Use menu > Algebra > Solve and complete as shown. Convert to a decimal answer using ctrl enter or menu > Actions > Convert to Decimal. Round to three decimal places as required: = Using the Casio ClassPad InM, enter and highlight the equation 3 2 = 28. Go to Interactive > Equation/Inequalit > solve and tap K. Cop and paste the answer into the net entr line and go to Interactive > Transformation > simplif to obtain a simplified eact answer. Highlight the answer and tap u to obtain the decimal approimation.

30 238 Chapter 5: Eponential and logarithmic functions 5G Eample 27 Solve the inequalit Taking log 0 of both sides: log 0 (0.7 log log log log log (direction of inequalit reversed since log < 0 Alternativel, we can solve the inequalit directl as follows: Note that 0 < 0.7 < and thus = 0.7 is strictl decreasing. Therefore the inequalit holds for log Section summar If a R + \{} and R, then the statements a = b and log a b = are equivalent. This defining propert of logarithms ma be used in the solution of eponential equations and inequalities. For eample: 2 = 5 = log log = 5 = log log An eponential inequalit ma also be solved b taking log a of both sides. For a >, the direction of the inequalit stas the same (as = log a is strictl increasing. For 0 < a <, the direction of the inequalit reverses (as = log a is strictl decreasing. Eercise 5G Eample 24 a If log 2 8 = k log , find the value of k. b If log 2 7 log 2 7 = 4, find the value of. c If log e 7 log e 4 =, find the value of. Eample 25, 26 2 Use our calculator to solve each of the following equations, correct to two decimal places: a 2 = 6 b 3 = 0.7 c 3 = d 4 = 5 e 2 = 5 f 0.2 = 3 g 5 = 3 h 8 = i 3 = 8 j = 0.7 k 2 = 3 + l.4 +2 = 25(0.9 m 5 = n 2 2 (+2 = 3 o = 00 3 Solve for using a calculator. Epress our answer correct to two decimal places. a 2 < 7 b 3 > 6 c 0.2 > 3 d e

31 5G 5H Inverses Solve each of the following equations for. Give eact answers. a 2 = 5 b 3 2 = 8 c 7 3+ = 20 d 3 = 7 e 3 = 6 f 5 = 6 g = 0 h = 0 Eample 27 5 Solve each of the following inequalities for. Give eact answers. a 7 > 52 b 3 2 < 40 c d e 3 < 06 f 5 < a If a log 2 7 = 3 log 6 4, find the value of a, correct to three significant figures. b If log 3 8 = log k, find the value of k, correct to one decimal place. 7 Prove that if log r p = q and log q r = p, then log q p = pq. 8 If u = log 9, find in terms of u: a b log 9 (3 c log 8 9 Solve the equation log 5 = 6 log 5. 0 Given that q p = 25, find log 5 q in terms of p. 5H Inverses We have observed that f ( = log a and g( = a are inverse functions. In this section, this observation is used to find inverses of related functions and to transform equations. An important consequence is the following: log a (a = for all R a log a = for all R + Eample 28 Find the inverse of the function f : R R, f ( = e + 2 and state the domain and range of the inverse function. Recall that the transformation reflection in the line = is given b the mapping (, (,. Consider = e = e = log e ( 2 Thus the inverse function has rule f ( = log e ( 2. Domain of f = range of f = (2,. Range of f = domain of f = R.

32 240 Chapter 5: Eponential and logarithmic functions Eample 29 Rewrite the equation = 2 log e ( + 3 with as the subject. 3 2 = 2 log e ( + 3 = log e = e 3 2 Eample 30 Find the inverse of the function f : (, R, f ( = 2 log e ( + 3. State the domain and range of the inverse. Solve = 2 log e ( + 3 for : 3 = log 2 e ( = e 3 2 = e Hence f ( = e Domain of f = range of f = R. Range of f = domain of f = (,. Using the TI-Nspire Use solve from the Algebra menu as shown. Using the Casio ClassPad Enter and highlight = 2 ln( + 3. Select Interactive > Equation/Inequalit > solve and ensure the variable is set to.

33 5H 5H Inverses 24 Eample 3 Rewrite the equation P = Ae kt with t as the subject. P = Ae kt Take logarithms with base e of both sides: log e P = log e (Ae kt = log e A + log e (e kt = log e A + kt t = ( loge P log k e A = ( P k log e A Section summar Let a R + \{}. The functions f : R R, f ( = a and g: R + R, g( = log a are inverse functions. That is, g = f. log a (a = for all a log a = for all positive values of Eercise 5H Eample 28 Find the inverse of the function f : R R, f ( = e 2 and state the domain and range of the inverse function. 2 n the one set of aes, sketch the graphs of = f ( and = f (, where f : R R, f ( = e n the one set of aes, sketch the graphs of = f ( and = f (, where f : (, R, f ( = log e (. Eample 29 Eample Rewrite the equation = 3 log e ( 4 with as the subject. Find the inverse of each of the following functions and state the domain and range in each case: a f : R + R, f ( = log e (2 b f : R + R, f ( = 3 log e (2 + c f : R R, f ( = e + 2 d f : R R, f ( = e +2 e f :( 2, R, f ( = log e (2 + f f :( 2 3, R, f ( = 4 log e (3 + 2 g f :(, R, f ( = log 0 ( + h f : R R, f ( = 2e

34 242 Chapter 5: Eponential and logarithmic functions 5H 6 The function f has the rule f ( = e. a Sketch the graph of f. b Find the domain of f and find f (. c Sketch the graph of f on the same set of aes as the graph of f. 7 Let f : R R where f ( = 5e 2 3. a Sketch the graph of f. b Find the inverse function f. c Sketch the graph of f on the same set of aes as the graph of f. 8 Let f : R + R where f ( = 2 log e ( +. a Sketch the graph of f. b Find the inverse function f and state the range. c Sketch the graph of f on the same set of aes as the graph of f. Eample 3 9 Rewrite the equation P = Ae kt + b with t as the subject. 0 For each of the following formulas, make the pronumeral in brackets the subject: a = 2 log e ( + 5 ( b P = Ae 6 ( c = a n (n d = 5 0 ( e = 5 3 log e (2 ( f = 6 2n (n g = log e (2 ( h = 5( e ( For f : R R, f ( = 2e 4: a Find the inverse function f. b Find the coordinates of the points of intersection of the graphs of = f ( and = f (. 2 For f :( 3, R, f ( = 2 log e ( : a Find the inverse function f. b Find the coordinates of the points of intersection of the graphs of = f ( and = f (. 3 a Using a calculator, for each of the following plot the graphs of = f ( and = g(, together with the line =, on the one set of aes: i f ( = log e and g( = e ii f ( = 2 log e ( + 3 and g( = e 3 2 iii f ( = log 0 and g( = 0 b Use our answers to part a to comment on the relationship in general between f ( = a log b ( + c and g( = b c a.

35 5I Eponential growth and deca 243 5I Eponential growth and deca We will show in Chapter that, if the rate at which a quantit increases or decreases is proportional to its current value, then the quantit obes the law of eponential change. Let A be the quantit at time t. Then A = A 0 e kt where A 0 is the initial quantit and k is the rate constant. If k > 0, the model represents growth: growth of cells population growth continuousl compounded interest If k < 0, the model represents deca: radioactive deca cooling of materials An equivalent wa to write this model is as A = A 0 b t, where we take b = e k. In this form, growth corresponds to b > and deca corresponds to b <. Cell growth Suppose a particular tpe of bacteria cell divides into two new cells ever T D minutes. Let N 0 be the initial number of cells of this tpe. After t minutes the number of cells, N, isgivenb N = N 0 2 t T D where T D is called the generation time. Eample 32 What is the generation time of a bacterial population that increases from 5000 cells to cells in four hours of growth? In this eample, N 0 = 5000 and N = when t = 240. Hence = T D 20 = T D Thus T D = (correct to two decimal places. log 2 20 The generation time is approimatel minutes. Radioactive deca Radioactive materials deca such that the amount of radioactive material, A, present at time t (in ears is given b A = A 0 e kt where A 0 is the initial amount and k is a positive constant that depends on the tpe of material. A radioactive substance is often described in terms of its half-life, which is the time required for half the material to deca.

36 244 Chapter 5: Eponential and logarithmic functions Eample 33 After 000 ears, a sample of radium-226 has decaed to 64.7% of its original mass. Find the half-life of radium-226. We use the formula A = A 0 e kt. When t = 000, A = 0.647A 0. Thus 0.647A 0 = A 0 e 000k = e 000k 000k = log e k = log e To find the half-life, we consider when A = 2 A 0: A 0 e kt = 2 A 0 e kt = 2 kt = log e ( 2 t = log e ( k The half-life of radium-226 is approimatel 592 ears. Population growth It is sometimes possible to model population growth through eponential models. Eample 34 The population of a town was 8000 at the beginning of 2007 and at the end of 204. Assume that the growth is eponential. a Find the population at the end of 206. b In what ear will the population be double that of 204? Let P be the population at time t ears (measured from Januar Then P = 8000e kt At the end of 204, t = 8 and P = Therefore = 8000e 8k 5 8 = e8k k = ( 5 8 log e The rate of increase is 7.9% per annum. Note: The approimation was not used in the calculations which follow. The value for k was held in the calculator.

37 5I Eponential growth and deca 245 a When t = 0, P = 8000e 0k The population is approimatel b When does P = ? Consider the equation = 8000e kt = ekt 5 4 = ekt t = ( 5 k log e The population reaches approimatel 6.82 ears after the beginning of 2007, i.e. during the ear Eample 35 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. Let G 0 be the population of gre kangaroos at the start. Then the number of gre kangaroos after n ears is G = G 0 (. n, and the number of red kangaroos after n ears is R = 0G 0 (0.95 n. When the proportions are reversed: G = 0R G 0 (. n = 0 0G 0 (0.95 n (. n = 00(0.95 n Taking log e of both sides: log e ( (. n = log e ( 00(0.95 n n log e. = log e 00 + n log e 0.95 log n = e 00 log e. log e i.e. the proportions of the kangaroo populations will be reversed after 30 ears.

38 246 Chapter 5: Eponential and logarithmic functions 5I Section summar There are man situations in which a varing quantit can be modelled b an eponential function. Let A be the quantit at time t. Then A = A 0 e kt where A 0 is the initial quantit and k is a constant. Growth corresponds to k > 0, and deca corresponds to k < 0. Eercise 5I Skillsheet Eample 32 A population of 000 E. coli bacteria doubles ever 5 minutes. a Determine the formula for the number of bacteria at time t minutes. b How long will it take for the population to reach 0 000? (Give our answer to the nearest minute. 2 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 (in cm of the tree t ears after the beginning of this period. The diameter is 52 cm after ear, and 80 cm after 3 ears. Calculate the values of the constants d 0 and m. 3 The number of people, N, who have a particular disease at time t ears is given b N = N 0 e kt. a If the number is initiall and the number decreases b 20% each ear, find: i the value of N 0 ii the value of k. b How long does it take until onl 5000 people are infected? Eample 33 4 Polonium-20 is a radioactive substance. The deca of polonium-20 is described b the formula M = M 0 e kt, where M is the mass in grams of polonium-20 left after t das, and M 0 and k are constants. At t = 0, M = 0 g and at t = 40, M = 5g. a Find the values of M 0 and k. b What will be the mass of the polonium-20 after 70 das? c After how man das is the mass remaining 2 g? 5 A quantit A of radium at time t ears is given b A = A 0 e kt, where k is a positive constant and A 0 is the amount of radium at time t = 0. a Given that A = 2 A 0 when t = 690 ears, calculate k. b After how man ears does onl 20% of the original amount remain? Give our answer to the nearest ear. 6 The half-life of plutonium-239 is ears. If 20 grams are present now, how long will it take until onl 20% of the original sample remains? (Give our answer to the nearest ear.

39 5I 5I Eponential growth and deca Carbon-4 is a radioactive substance with a half-life of 5730 ears. It is used to determine the age of ancient objects. A Bablonian cloth fragment now has 40% of the carbon-4 that it contained originall. How old is the fragment of cloth? Eample 34 Eample The population of a town was at the beginning of 2002 and at the end of 204. Assume that the growth is eponential. a Find the population at the end of 207. b In what ear will the population be double that of 204? There are approimatel five times as man magpies as currawongs in a certain area. If the population of currawongs increases at a rate of 2% per annum while that of the magpies decreases at 6% per annum, find how man ears must elapse before the proportions are reversed, assuming the same rates continue to appl. 0 The pressure in the Earth s atmosphere decreases eponentiall as ou rise above the surface. The pressure in millibars at a height of h kilometres is given approimatel b the function P(h = h. a Find the pressure at a height of 4 km. (Give our answer to the nearest millibar. b Find the height at which the pressure is 450 millibars. (Give our answer to the nearest metre. A biological culture contains bacteria at 2 p.m. on Sunda. The culture increases b 0% ever hour. At what time will the culture eceed 4 million bacteria? 2 When a liquid is placed into a refrigerator, its temperature T C at time t minutes is given b the formula T = T 0 e kt. The temperature is initiall 00 C and drops to 40 C in 5 minutes. Find the temperature of the liquid after 5 minutes. 3 The number of bacteria in a certain culture at time t weeks is given b the rule N = N 0 e kt. If when t = 2, N = 0 and when t = 4, N = 203, calculate the values of N 0 and k. 4 Five kilograms of sugar is graduall dissolved in a vat of water. After t hours, the amount, S kg, of undissolved sugar remaining is given b S = 5 e kt. a Calculate k given that S = 3.2 when t = 2. b At what time will there be kg of sugar remaining? 5 The number of bacteria, N, in a culture increases eponentiall with time according to the rule N = a b t, where time t is measured in hours. When observation started, there were 000 bacteria, and 5 hours later there were bacteria. a Find the values of a and b. b Find, to the nearest minute, when there were 5000 bacteria. c Find, to the nearest minute, when the number of bacteria first eceeds d How man bacteria would there be 2 hours after the first observation?

40 248 Chapter 5: Eponential and logarithmic functions Review Spreadsheet AS Chapter summar Sketch graphs of the form = a and transformations of these graphs. = 0 = e = 2 Nrich (0, Inde laws a m a n = a m+n a m a n = a m n (a m n = a mn Logarithms For a R + \{}, the logarithm function with base a is defined as follows: a = is equivalent to log a = Sketch graphs of the form = log a and transformations of these graphs. = log 2 = log e = log 0 (, 0 Logarithm laws ( m log a (mn = log a m + log a n log a = log n a m log a n ( log a = log n a n log a (m p = p log a m Change of base log a = log b log b a and a = b (log b a Inverse functions The inverse function of f : R R, f ( = a is f : R + R, f ( = log a. log a (a = for all R a log a = for all R + Law of eponential change Assume that the rate at which the quantit A increases or decreases is proportional to its current value. Then the value of A at time t is given b A = A 0 e kt where A 0 is the initial quantit and k is a constant. Growth corresponds to k > 0, and deca corresponds to k < 0.