math analysis Ch 3 review

Review problems for test:Part 1: non-calculator partDescribe the transformation:1. f(x) = e x , g(x) = -2e x – 3 + 42. f(x) = 4 x , g(x) = 4 -2x – 83. f(x) = e x , g(x) = e x + 3 – 3Solve4. log 1000 = x5. log 3 = x6. log 4 1/64 = x7. log x = 28. log 5 x = 39. ln e 4 = x10. = x11. = x12. = xUsing properties of logs, expand the following functions13. log 8x14. ln 4/x15. log 2 x 316. lnUsing properties of logs, condense the following17. log 4 + log 618. log x – log 419. 4ln x + 6 ln yPart 2: calculator partUse a calculator to evaluate:20. log 521. log -322. log 3 + 523. ln 824. log 3 725. log 2 926. The population in a town in 2000 was 24,750. If the population is increasing at a rate of2.5% per year find the following:a. Write the exponential functionb. What would the population be in 2004?c. In what year will the population reach 40,00027. There is 6.5 g of a certain substance has a half-life of 35 days.a. Write the exponential function in terms of t.b. How much is left after 10 days,c. When will there be less than 5 grams remaining?

28. Determine the atmospheric pressure outside an aircraft flying at 8.5 miles above sealevel?Solve the following equations. Round to the nearest thousandth29. 3 2x = 2730. 2(1/4) 4x = 3231. 5 x – 4 = 1032. 6 -2x = 1833. log 2 32 = x34. log 4 x-4 = 435. 4 + 3 ln (x – 3) = 836. -2 ln 4x = 537. e -2x = 1038. 4 - 2e x = 339. You deposit $400 into an account that pays 3.5% simple interest. After 5 years, howmuch will you have?40. You deposit $4500 into an account that pays 2.75% interest compounded quarterly.After 3 years, how much will you have?41. You deposit $1000 into an account that pays 2.4% interest compounded continuously,have much will you have after 8 years?42. What is the interest rate compounded monthly is required for $500 to reach $800 in 5years?43. Find the APY for an investment with an interest rate of 8.25% compounded monthly.44. Find the future value of a retirement account in which Zachary makes payments of $150in which interest is credited quarterly with an interest rate of 6.5% after 25 years.45. Find the present value of a loan in which payments of $247.60 are made quarterly withan interest rate of 3.25% for 6 years.46. How much are the monthly payments on a loan in which $14,500 is borrowed at a rate of2.25% for 3 years.47. You contribute $50 per month into an account that earns 4.5% annual interest. After 30years, how much will you have?

Review problems for test:Part 1: non-calculator partANSWERS:Describe the transformation:1. f(x) = e x , g(x) = -2e x – 3 + 4: reflection over the x-axis, vertical stretch, shifted right 3 andup 42. f(x) = 4 x , g(x) = 4 -2x – 8: reflection over the y-axis, horizontal shrink, shifted down 83. f(x) = e x , g(x) = e x + 3 – 3: shifted left 3 and down 3Solve4. log 1000 = x: 10 x = 1000, x = 35. log 3 = x: log3 81 1/4 = x, 3 x = 81 1/4 , 3 x = 3 4(1/4) , x = 4(1/4), x = 16. log 4 1/64 = x: log4 64 -1 = x, 4 x = 64 -1 , 4 x = 4 3(-1) , x = 3(-1), x = -37. log x = 2:10 2 = x, 100 = x8. log 5 x = 3:5 3 = x, 125 = x9. ln e 4 = x: 4 = x (remember ln and e cancel)10. e ln 4 = x: 4 = x11. = x: 8 = x (remember 4 and log 4 cancel)12. = x:4 = xUsing properties of logs, expand the following functions13. log 8x: log 8 + log x14. ln 4/x: ln 4 – ln x15. log 2 x 3 : 3 log 2 x16. ln : ln x 1/5 – ln y 1/3 = 1/5 ln x – 1/3 ln yUsing properties of logs, condense the following17. log 4 + log 6: log 4(6) = log 2418. log x – log 4: log x/419. 4ln x + 6 ln y: ln x 4 + ln y 6 = ln x 4 y 6Part 2: calculator partUse a calculator to evaluate:20. log 5: .69921. log -3: undefined22. log 3 + 5: 5.47723. ln 8: 2.07924. log 3 7: 1.77125. log 2 9: 3.17026. The population in a town in 2000 was 24,750. If the population is increasing at a rate of2.5% per year find the following:a. Write the exponential function: P(t) = 24,750(1 + .025) tb. What would the population be in 2004? P(4) = 24,750 (1.025) 4 = 27,319 or 27,320

c. In what year will the population reach 40,000: 2020, (enter the function into thecalculator as a graph and go to the table of values, scroll until y is 40,000 or more)27. There is 6.5 g of a certain substance has a half-life of 35 days.a. Write the exponential function in terms of t.: A(t) = 6.5(1/2) t/35b. How much is left after 10 days: 5.332 gc. When will there be less than 5 grams remaining?14 days28. Determine the atmospheric pressure outside an aircraft flying at 8.5 miles above sealevel? P(h) = 14.7(1/2) 8.5/3.6 = 2.861 pounds per sq. inch (remember it is 14.7 and theexponent is h/3.6)Solve the following equations. Round to the nearest thousandth29. 3 2x = 27: 3 2x = 3 3 , 2x = 3, x = 3/230. 2(1/4) 4x = 32: (1/4) 4x = 16, 4 -4x = 16, 4 -4x = 4 2 , -4x = 2, x = -1/231. 5 x – 4 = 10: 5 x = 14, log 5 5 x = log 5 14, x = log 14/log 5, x = 1.640 (use change-of-baseformula)32. 6 -2x = 18: log 6 6 -2x = log 6 18, -2x = log 18/log 6, -2x = 1.6131…, x = -.80733. log 2 32 = x: 2 x = 32, 2 x = 2 5 , x = 534. log 4 x-4 = 4: 4 4 = x- 4, 256 = x – 4, 260 = x35. 4 + 3 ln (x – 3) = 8: 3ln (x – 3) = 4, ln (x – 3) = 4/3, e ln(x-3) = e 4/3 , x – 3 = 3.7936…,x = 6.79436. -2 ln 4x = 5: ln 4x = -5/2, e ln 4x = e -5/2 , 4x = .08208…, x = .020537. e -2x = 10: ln e -2x = ln 10, -2x = 2.3025…, x = -1.15138. 4 - 2e x = 3: -2e x = -1, e x = ½, ln e x = ln ½, x = -.69339. You deposit $400 into an account that pays 3.5% simple interest. After 5 years, howmuch will you have?A = 400(1 + .035) 5 = $475.0740. You deposit $4500 into an account that pays 2.75% interest compounded quarterly.After 3 years, how much will you have?A = 4500(1 + .0275/4) 4(3) = $4885.6141. You deposit $1000 into an account that pays 2.4% interest compounded continuously,have much will you have after 8 years?A = 1000e .024(8) = $1211.6742. What is the interest rate compounded monthly is required for $500 to reach $800 in 5years?800 = 500(1 + r/12) 12(5) , 1.6 = (1 + r/12) 60 , 1.00786… = 1 + r/12, .0078… = r/12, .0943…= r, 9.43% = r

43. Find the APY for an investment with an interest rate of 8.25% compounded monthly.(1 + x) = (1 + .0825/12) 12 , 1 + x = 1.0856…, x = .0856.., 8.57% is the APY44. Find the future value of a retirement account in which Zachary makes payments of $150in which interest is credited quarterly with an interest rate of 6.5% after 25 years.FV = 150 (1 + .065/4) 4(25) – 1) , (1+.065 ÷ 4)^100 enter, -1 enter, ÷ (.065÷4) enter, ∙ 150.065/4 FV = $37,038.6245. Find the present value of a loan in which payments of $247.60 are made quarterly withan interest rate of 3.25% for 6 years.PV = 247.60 1 – (1 + .0325/4) -4(6) , (1 + .0325 ÷ 4)^-24 enter, 1 – 2 nd (-) enter,.0325/4 ÷(.0325÷4) enter, ∙ 247.60 enter, PV = $5379.1446. How much are the monthly payments on a loan in which $14,500 is borrowed at a rate of2.25% for 3 years.14,500(.0225/12) = R (1 – (1 + ) -12(3) , (1 + .0225÷ 12)^ -36 enter, 1 – 2 nd (-) enter,14,500(.0225÷12) ÷ 2 nd (-) enterPayment: $416.9047. You contribute $50 per month into an account that earns 4.5% annual interest. After 30years, how much will you have?FV = 50 (1 + .045/12) 12(30) – 1 , FV = $37,969.31.045/12