Chapter 6. Exponential and Logarithmic Functions

Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.1 1) (f + g)(x) = f(x) + g(x) = (2x+3) + (3-x) = 2x+3 + 3 – x = x + 6 The function is defined for all values of x, so the domain is all real numbers. Solution: (f + g)(x)= x + 6 3) (f/g)(x) To find the domain, ignore the numerator. Find where the denominator equals zero. Exclude this number. domain : computation 3 – x = 0 3=x Solution: (f/g)(x) = 7) 11) (g – f)(x) ( ) The function is defined for all values of x, so the domain is all real numbers. The function is defined for all values of x, so the domain is all real numbers. Solution: Solution: (g – f)(x) = -2x2 + 10x + 7 = -2x 13) 15) ( =10x3-17x2 -35x - 12 The function is defined for all values of x, so the domain is all real numbers. Solution: = 10x3 – 17x2 – 35x – 12 ) =2(25x2 + 20x + 20x + 16) – 25x – 20 – 3 = 50x2 + 40x + 40x + 32 – 25x – 20 – 3 =50x2 +55x+9 The function is defined for all values of x, so the domain is all real numbers. Solution: = 50x2 + 55x + 9 117 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.1 19) (h+k)(3) [ 21) (h/k)(5) ] Solution: (h+k)(3)=9 Solution: (h/k)(5) = 25) 27) ( ) ( Solution: =1 ) Solution: = -6 35) 31) ( ) ( ) Solution: =9 Solution: ( ) 37) 41) (fg)(0) ( ) To find f(0) look at the graph. This is asking for the y coordinate of the point that has 0 for its x coordinate (f(0) = 1) Solution: = 99 To find g(0) look at the graph. This is asking for the y coordinate of the point that has 0 for its x coordinate. (g(0) = -1) Solution: (fg)(0)=-1 118 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.1 43) (g/f)(1) 47) First write without composite symbol. =g(f(-2)) **** To find f(1) look at the graph. This is asking for the y coordinate of the point that has 0 for its x coordinate. ( f(1) = 3) To find g(1) look at the graph. This is asking for the y coordinate of the point that has 0 for its x coordinate. ( g(1) = 0) Then find f(-2), which is the y coordinate of the point on the f graph with x of -2. f(-2)=-3 replace f(-2) with 3 in **** above. =g(-3) Solution: (g/f)(1)= 0 Now find g(-3) which is the y coordinate of the point on the g graph with x = -3. g(-3) =8 Solution: =8 _____________________________________________________________________________________ Section 6.2 #5-12, Sketch a graph and determine whether each function is one to one (you may construct a table of values, use a graphing calculator, or use a technique you already have learned to construct your graph.) 5) f(x) = 2x – 5 7) f(x) = x2 – 3 f(x) = 2x – 5 y y 5 5 x x -8 -8 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 8 8 -5 -5 Function is one-to-one no horizontal line can be drawn to touch the graph in more than one place, the function is one-to-one. Function is not one-to-one A horizontal line can be drawn to touch the graph in more than one place, the function not is one-toone. 119 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.2 9) g(x) = x4 11) f(x) = x3 y y 5 5 x x -8 -6 -4 -2 2 4 6 -8 -6 -4 -2 2 4 6 8 8 -5 -5 Solution: Function is one-to-one Solution: Function is not one-to-one A horizontal line can be drawn to touch the graph in more than one place, the function not is one-toone. No horizontal line can be drawn to touch the graph in more than one place, the function is one-to-one. 13) f = { (0,1) (1,4) (2,4) (3,5)} 15) h = {(0,3) (5,1) (7,11) (9, -3)} The function is not one-to-one because two different points have the same y-value. Function is one-to-one because all of the y values are different. Find inverse by switching the x and y’s. Solution: Function is not one-to-one Solution: Function is one-to-one h-1 = { (3,0) (1,5) (11,7) (-3,9)} 120 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.2 19) f(x) = 2x – 4 first: replace function symbol with a y y = 2x - 4 second: switch x and y, this creates the inverse x = 2y - 4 third: solve for y 21) first: replace function symbol with a y second: switch x and y, this creates the inverse third: solve for y fourth: write with y on left side (if not already there) fifth: replace the y with an inverse symbol Solution: 25) √ first: replace function symbol with a y √ second: switch x and y, this creates the inverse √ third: solve for y (√ ) x3 = y fourth: write with y on left side (if not already there) y = x3 fifth: replace the y with an inverse symbol Solution: 3x = y – 2 3x+ 2 = y fourth: write with y on left side (if not already there) y = 3x + 2 fifth: replace the y with an inverse symbol Solution: 27) f(x) = x3 + 2 first: replace function symbol with a y y = x3 + 2 second: switch x and y, this creates the inverse x=y3+2 third: solve for y x-2 = y3 √ √ √ fourth: write with y on left side (if not already there) √ fifth: replace the y with an inverse symbol Solution: √ 121 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.3 13) f(x) = 3x ( ) 15) y y 5 5 x -8 -6 -4 -2 2 4 6 x 8 -8 -6 -4 -2 2 4 6 8 -5 -5 x 0 1 -1 -2 f(x) 1 3 1/3 1/9 computations f(0)=30 = 1 f(1)=31 = 3 f(-1) = 3-1 = 1/3 f(-2)=3-2 = 1/32 = 1/9 x 0 h(x) computations 1 -1 1/3 1 3 -2 9 2 1/9 -3 27 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 122 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.3 17) 19) y y 5 5 x x -8 -6 -4 -2 2 4 6 -8 8 -6 f(x) 1 2 ½ 4 1/4 computations f(3) = 23-3 = 20 = 1 f(4) = 24-3 = 21 = 2 f(2)=22-3 = 2-1 = 1/2 f(5)= 25-3 = 22 = 4 f(1)=21-3 = 2-2 = ½2 = 1/4 -2 2 4 6 8 -5 -5 x 3 4 2 5 1 -4 x -4 -5 -3 -6 f(x) 1 ½ 2 1/4 computations f(-4)=2-4+4 = 20 = 1 f(-5) = 2-5+4 = 2-1 = 1/2 f(-3) = 2-3+4 = 21 = 2 f(-6) = 2-6+4 = 2-2 = ½2 = 1/4 23) The number of computers infected by the spread of a virus through email can be described by the exponential function c(t)=4(1.02)t , where t is the number of minutes since the first infected e-mail was opened. Approximate the number of computers that will be infected after 6 hours (240 minutes). (round to the nearest computer) I just need to plug in 240 for t in the equation. I will need to use my calculator to get this answer. c(240) = 4(1.02)240 = 464 Solution: 464 computers 25) The charge remaining in a battery decreases as the battery discharges. The charge C (in coulombs) after t days is given by the formula C(t)= 0.0003(0.7)t. Find the charge after 5 days. (carry 6 decimal places in your answer) I need to evaluate the function at t = 5. I will need my calculator to simplify. C(0) = 0.0003(0.7)5 = 0.00005 Solution: 0.00005 coulombs 123 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.3 #26-30: Use the compound interest formula ( ) to answer the following. 27) An initial deposit of $1,000 earns 4% interest compounded twice per year. How much will be in the account after 5 years? P = 1,000 r = .04 n=2 t=5 Substitute these values in the formula. Then use my calculator to evaluate. ( ) =1218.99 Solution: $1218.99 124 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.4 1) 32 = 9 5) The base of the exponential function (3) is the base or subscripted part of the logarithm. Switch the 2 and the 9. The base of the exponential function (3) is the base or subscripted part of the logarithm. Switch the -1 and the 1/3. Solution: log39 =2 Solution: log3(1/3) =-1 7) ey = x 11) log381=4 The base of the exponential function (e) is the base or subscripted part of the logarithm. Switch the y and the x. Write the problem without the log. The 3 remains the base in the exponential function. Switch the 81 and 4. Solution: logex = y Solution: 34 =81 13) log264=6 17) log x=3 Write the problem without the log. The 2 remains the base in the exponential function. Switch the 64 and 6. Since no base is written the base is assumed to be 10. You can think of the problem like this: Solution: 26 = 64 log10x = 3 Write the problem without the log. The 10 remains the base in the exponential function. Switch the x and 3. Solution: 103 = x 125 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.4 19) log22 23) log31 This is asking the question 2? = 2, The answer must be 1 because: 21 = 2 This is asking the question 3? = 1, The answer must be 0 because: 30 = 1 Solution: 1 Solution: 0 27) log464 31) log1 This is asking the question 4? = 64, The answer must be 3 because: 43 = 64 Remember if no base is written the base is assumed to be a 10. Think of the problem as: Solution: 3 log101 This is asking the question 10? = 1, The answer must be 0 because: 100 = 1 Solution: 0 33) log100 37) log223 This is asking the question 10? = 100, The answer must be 2 because: 102 = 100 This is asking the question 2? = 23, The answer must be 3 because: 23 = 23 Solution: 2 Solution: 3 41) log445 This is asking the question 4? = 45, The answer must be 5 because: 45 = 45 Solution: 5 126 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.4 49) y = log2x 2y= x (problem written in exponential form) Domain x y 1 0 computations (y=0) 20 = x 1=x (y = -1) 2-1 = x ½=x (y=1) 21 = x 2=x (y=-2) 2-2 = x ¼=x (y=2) 22 = x 4=x ½ -1 2 1 ¼ -2 4 2 y 5 x -8 -6 -4 -2 2 4 6 8 -5 53) y = log2(x+4) 2y = x + 4 (problem written in exponential form) Domain x -3 -3.5 y 0 -1 -2 1 -3.75 -2 0 2 computations (y = 0) 20 = x+ 4 1=x+4 -3 = x (y=-1) 2-1 = x ½=x+4 .5 = x+4 -3.5 = x (y =1) 21 = x+4 2 =x+4 -2 = x (y = -2) 2-2 = x+4 ¼ = x+4 0.25 = x+4 -3.75 = x (y=2) 22 = x+4 4 = x+4 0=x y 5 x -8 -6 -4 -2 2 4 6 8 -5 127 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.4 57) 13) y= ex (from 6.5) ⁄ y equation written in exponential form (1/2)y = x + 1 Domain 5 y x 5 -8 -6 -4 -2 2 4 6 8 x -8 -6 -4 -2 2 4 6 -5 8 -5 x 0 1 y 0 -1 -1/2 1 3 -2 -3/4 2 computations (y = 0) (1/2)0 = x+1 1 = x+1 0=x (y = -1) (1/2)-1 = x+ 1 (2/1)1 = x+ 1 2 = x+1 1=x (y=1) (1/2)1 = x+1 ½ = x+1 ½-1 = x x Y 0 1 -1 .37 1 2.72 -2 .14 2 7.39 computations – done on calculator y=e0 y=1 y=e-1 =.37 y = e1 y = 2.72 y = e-2 y = .14 y = e2 y=7.39 (y = -2) (1/2)-2 = x+1 (2/1)2 = x+1 4 =x+1 3=x (y=2) (1/2)2 = x+1 ¼ = x+1 ¼-1=x -3/4 = x 128 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.5 17) y = ex-2 19) y = ln(x) Equation written in exponential form: ey = x Domain y 5 y x -8 -6 -4 -2 2 4 6 5 8 x -5 -8 -6 -4 -2 2 4 6 8 -5 x Y 2 0 3 2.72 1 .37 4 7.39 0 .14 computations – done on calculator y = e2-2 y= e0 y=1 y = e3-2 y = e1 y = 2.72 y= e1-2 y = e-1 y = 0.37 y= e4-2 y = e2 y = 7.39 y=e0-2 y = e-2 y = 0.14 x 1 y 0 2.72 1 .37 -1 7.39 2 .14 -2 computations- done on calculator (y=0) e0 = x 1=x (y = 1) e1 = x 2.72 = x (y = -1) e-1 = x 0.37 = x (y = 2) e2 = x 7.39 = x (y = -2) e-2 = x 0.14 = x 129 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.5 23) y = ln(x+2) Equation written in exponential form: ey = x+2 Domain 25) y = ln(3 – x) Equation written in exponential form: ey = 3 – x I will rewrite a bit more x = 3 – ey Domain y 5 y x -8 -6 -4 -2 2 4 6 5 8 x -5 -8 -6 -4 -2 2 4 6 8 -5 x -1 y 0 .72 1 -1.63 -1 5.39 2 -1.86 -2 computations, done on calculator (y=0) e0 = x+2 1 = x+2 -1 = x (y=1) e1 = x+2 2.72 = x+2 x = .72 (y=-1) e-1 = x+ 2. .37 = x+ 2 -1.63 = x (y = 2) e2 = x+2 7.39 = x+2 5.39 =x (y=-2) e-2 = x+2 0.14 = x+2 -1.86 = x x 2 2.63 .28 2.86 -4.39 y 0 -1 1 -2 2 computations- done on calculator (y=0) x = 3-e0 = 2 (y=-1) x = 3-e-1 = 2.63 (y=1) x = 3-e1 = 3-2.72 = .28 (y=-2) x = 3-e-2 = 3-.14 = 2.86 (y=2) x = 3-e2 = 3 – 7.39 = -4.39 130 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.5 29) An initial investment of $10,000 earns 5.25% interest compounded continuously. What will the investment be worth in 8 years? P = 10000 r = .0525 t=8 Substitute values in formula Use a calculator to evaluate. A=10000e8*.0525 = 15219.62 Solution: $15219.62 _____________________________________________________________________________________ Section 6.6 1) log216 5) log885 This is asking the question 2? = 16, the answer is 4, because 24 = 16 This is asking the question 8? = 85, the answer is 5, because 85 = 85 Solution: 4 Solution: 5 9) ln(e) Think of this as logee This is asking the question e? = e, the answer is 1, because e1 = e 11) log2643 This is asking the question 2? = 643 I will need to do a bit of algebra to solve this. I know 26 = 64. I can rewrite the problem. 2? = (26)3 I will clear the parenthesis by multiplying exponents 2? = 218 Now it should be obvious that 18 is the solution. Solution: 18 Solution: 1 Section 6.6 13) Which of these is a true statement? a) If I replace with the values to the left: b) If I replace with the values above: 131 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems a) log2(4*8) = log24 * log28 b) log2(4*8) = log24 + log28 To answer these questions I need to know what each part equals. log2(4*8) = log24 * log28 log2(4*8) = log24 + log28 5=2+3 5 = 5 (this is true 5 = 2*3 5=6 This is false. log2(4*8) = log232 = 5 log24 = 2 log2 8 = 3 Solution: a is false, b is true 15) Which of these is a true statement? a) If I replace with the values above: b) If I replace with the values above: a) b) (this is false) 3=4-1 3=3 (this is true) I need to know what each piece is equal to, before I can answer the question. log216 = 4 log22 =1 Solution: a is false, b is true Section 6.6 17) Which of these is a true statement? a) log1003 = (log100)3 b) log1003 = = 3log100 a) I will replace the values to determine if this is true or not. b) I will replace the values to determine if this is true or not. log1003 = (log100)3 log1003 = = 3log100 132 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems I need to compute each piece before I can answer the question. You can use your calculator if you can’t do these by hand yet. 6 = (2)3 6 = 8 (this is false) 6 = 3*2 6 = 6 (this is true) log1003 = 6 log100 = 2 Solution: a is false, b is true 21) log5(25x2y6) 23) = log525 + log5x2 + log5y6 = 2 + 2log5x + 6log5y =log2(xy3) – log2z2 = log2x + log2y3 – log2z2 Solution: log525 + 2log5x + 6log5y or 2 + 2log5x + 6log5y = log2x + 3log2y – 2log2z Solution: log2x + 3log2y – 2log2z 25) 29) = log2(xy) – log2(w2z5) = = log2x + log2y – (log2w2 + log2z5) = log2x + log2y – 2log2w – 5log2z Solution: log2x + log2y – 2log2w – 5log2z √ = √ ⁄ = 2log2x + log2y Solution: 2log2x + log2y Section 6.6 33) 2log3x + 4log3y + log3z 35) 5log2x + 3log2y – log2z = log3x2 + log3y4 + log3z = log2x5 + log2y3 – log2z 133 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems = log3x3y4z =log2(x5y3) – log2z Solution: log3x3y4z = Solution: 37) 4logx – 2logy – 3 logz =logx4 – (2logy + 3 logz) = logx4 – (logy2 + logz3) =logx4 – log(y2z3) = Solution: 134 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.7 3) 2x+1 = 32 2x+1 = 25 5) ( ) x+1 = 5 I need to use this fact to solve this problem without logarithms. x=4 This solution will check, but I am not showing the work. Solution: x = 4 (2-1)x = 24 2-x = 24 -x = 4 x = -4 Solution: x = -4 Check x = -4 ( ) ( ) Th e solution checks. x = -4 is the only solution 7) 24-x = 26 4 – x =6 -x = 6 – 4 -x = 2 x = -2 Solution: x = -2 Check x = -2 24- (-2) = 64 24+2 = 64 26 = 64 64 = 64 9) 32x = 2 (25)x = 21 25x = 21 5x = 1 x = 1/5 Solution: x = 1/5 Check x = 1/5 ⁄ √ The solution checks. x = 1/5 is the only solution. The solution checks. x = -2 is the only solution. 135 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.7 13) 3x = 6 take log of both sides: log3x = log6 make exponent coefficient: xlog3 = log 6 divide: x= ( I will get a decimal approximation using my calculator) Solution: x = log36 approx: 1.63 Check x = 1.63 (because I am checking a rounded answer, this might not check quite as nicely as the last few problems. 31.63 = 6 5.9938 = 6 (this isn’t exactly 6=6, but it is close enough) 15) ex = 12 I will take the ln of both sides, because of the e in the problems. take ln of both sides: ln(ex) = ln(12) make exponent coefficient: xlne = ln(12) substitute ln(e) =1: x(1) = ln(12) x = ln(12) x= 2.48 Solution: x = ln(12) approx: 2.48 Check x = 2.48 e2.48 = 12 11.941 = 12 (this is still close enough because I am using a rounded answer) 19) 32e2x = 128 first divide by 32 : e2x = 4 I will take the ln of both sides: ln(e2x) = ln4 make exponent coefficient: 2xlne = ln4 replace lne with 1: 2x = ln4 divide: (This solution will check, I am not going to show how this solution checks) Solution: approx: .69 136 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.7 25) log3x = 2 rewrite in exponential form: 32 = x 9=x Solution: x = 9 27) ln x = 1 rewrite with ln = loge logex = 1 Check x = 9 write without log log39 = 2 (you can evaluate the logarithm on your calculator, if you can’t do it in your head) 2=2 e1 = x The solution checks and x = 9 is the only solution. Solution: x = e (approx 2.72) check x = e ln e = 1 1=1 The solution checks. x = e is the only solution. 31) log4(3x-6) = -1 write in exponential form 4-1 = 3x-6 (cross multiply) 4(3x-6) = 1*1 12x – 24 = 1 12x = 25 x = 25/12 Solution: x = 25/12 check x = 4 log464 = 3 (again, use your calculator if you can’t do the log in your head) 3=3 The solution checks. x = 4 is the only solution. 35) log2(x-1) = 3 write without log 23 = x – 1 8=x–1 9=x Solution: x = 9 Check x = 9 log2(9-1) = 3 log28 = 3 (again, use your calculator if you can’t do the log in your head) 3=3 The solution checks. x = 3 is the only solution. 137 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.7 37) log2(2x)=5 write without the logarithm 5 2 = 2x 32 = 2x 16 = x (This solution checks, but I will not include the checking here.) Solution: x = 16 39) log(x+1) = log(3x-2) you can solve be dropping the logs and equating the arguments. x+1 = 3x -2 -2x = -3 = Solution: x = 3/2 Check x = 3/2 ( ) ( ( ) ( ( ) ) ) ( ) The solution checks. x=3/2 is the only solution. 138 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.7 43) log2x – log2(x+6)=-2 45) log2 x – log2 (x – 6) = 2 write as a single log write a single log: write without the log, in exponential form write in exponential form: replace 2-2 with 1/4: cross multiply 4x = x + 6 3x = 6 x=2 Solution: x = 2 Check x = 2 log22 - log2(2+6) = -2 1 – log28 = -2 1-3 = -2 -2 = -2 The solution checks. x = -2 is the only solution. clear fractions: 4x – 24 = x 3x = 24 x=8 Solution: x = 8 Check x = 8 log28 - log2(8-6) = 2 3 – log22 = 2 3–1=2 2=2 The solution checks. x= 8 is the only solution 139 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.7 47) log2(x+6) – log2(3x+2)=-1 51) log3(x+6) + log3(3x) = 4 write as a single log: write as a single log: write in exponential form: simplify: log3(3x2 + 18x) = 4 write in exponential form: 34 = 3x2 + 18x -1 replace 2 with 1/2": 81 = 3x2 + 18x cross multiply: 1(3x+2) = 2(x+6) divide by 3: Solve: 3x + 2 = 2x + 12 x = 10 This solution will check. I am not including the work. Solution: x = 10 27 = x2 + 6x 0 = x2 + 6x – 27 0 = (x+9)(x-3) x+9 = 0 or x – 3 =0 x = -9, or x =3 Solution: x =3 x = -9 is extraneous Check x = 3 Check x = -9 log3(3+6) + log3(3*3) = 4 log3(-9+6)+log3(3*(-9)) =4 log39 + log39 = 4 log3(-3) + log3(-27) = 4 2+2=4 Each of these logs is a non-real number. The solution checks. x = 3 is the only solution. The solution does not check x = -9 is extraneous 140 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.7 Use the formula: P = P0ekt , where P0 is the initial population at t=0, and k is the rate of growth. 57) The bacterial in a laboratory culture increased from an initial population of 500 to 1,500 in 3 hours. How long will it take the population to reach 10,000? (hint first use the 500 to 1,500 in 3 hours and the formula above to solve for k, then use the 10,000 to answer the question.) (round t the nearest hour) 57) continued: Use the first group of numbers and the formula to solve for k. P = 1500 P0 = 500 t=3 (make exponent a coefficient) Substitute values into equation: ⁄ put the values in the formula (divide by 500) ⁄ (take the ln of both sides) ⁄ (replace lne with 1) 1500 = 500ek*3 (multiply by (divide by 500) 3 = e3k (take the ln of both sides) ln(3) = ln(e3k) (make the exponent a coefficient) 3kln(e) (substitute lne = 1) ln(3) = (use your calculator) 8.1805 = t Solu Solution: About 8 hours ln(3) = 3k divide by 3 Now use the next bit of information to answer the question. P = 10000 P0 = 500 k= continued in right column 141 Chapter 6: Exponential and Logarithmic Functions Selected Solutions to Odd Problems Section 6.7 Use the continuous compound interest formula: A = Pert , where P is the initial investment, r the annual interest rate, and t the number of years to solve the following. 61) How long will it take an initial investment of $1,000 to double if it is expected to earn 6% interest compounded continuously? (round to 1 decimal place) 63) How long will it take an initial investment of $10,000 to grow to $15,000 if it is expected to earn 4% interest compounded continuously? (round to 1 decimal place) find values for the formula P = 1,000 A = 2,000 (goal is to double the initial investment. r = .06 find values for the formula: Substitute these values into the formula. 2000=1000e.06t A = 15000 P = 10000 r = .04 substitute into formula 15000= 10000e.04t (divide by 1000) 2 = e.06t (divide by 10000) 1.5 = e.04t .06t (take ln of both sides) ln(2) = ln(e ) (take ln of each side) ln(1.5) = ln(e.04t) (make exponent coefficient) ln(2) = (.06t)lne (make exponent a coefficient) ln(1.5) = .04tln(e) (lne = 1) ln(2) = .06t (ln(e) = 1) ln(1.5) = .04t divide: (divide) (use calculator) 11.6 = t Solution: about 11.6 years (use calculator) t = 10.1 Solution: about 10.1 years 142