Review problems for test:
Part 1: non-calculator part
Describe the transformation:
1. f(x) = ex, g(x) = -2ex – 3 + 4: reflection over the x-axis, vertical stretch of 2, shifted right 3, and up 4
2. f(x) = 4x, g(x) = 4-2x – 8: reflection over the y-axis, horizontal shrink of 2, and shifted down 8
3. f(x) = ex, g(x) = ex + 3 – 3: shift left 3 and down 3
Solve
4. log 1000 = x: 10x = 1000, 10x = 103, x = 3
5.
6.
7.
8.
9.
10.
11.
12.
log3
= x: log3
= x, 3x =
, 3x =
, 3 x = 3 1, x = 1
-1
x
-1
x
3(-1)
log4 1/64 = x: log4 64 = x, 4 = 64 , 4 = 4 , x = -3
log x = 2: 102 = x, 100 = x
log5 x = 3: 53 = x, 125 = x
ln e4 = x: 4 = x
= x: 4 = x
= x: 8 = x
= x: 4 = x
Using properties of logs, expand the following functions
13. log 8x: log 8 + log x
14. ln 4/x: ln 4 – ln x
15. log2 x3: 3 log2 x
16. ln
: 1/5 ln x – 1/3 ln y
Using properties of logs, condense the following
17. log 4 + log 6: log 24
18. log x – log 4: log x/4
19. 4ln x + 6 ln y: ln
Analyze the following function:
20. f(x) = 2(3)-x: d(-∞, ∞) r: (0, ∞) bounded, no max/min, continuous, decreasing: (-∞, ∞), neither, VA:
none, HA: none (technically y = 0), lim x -∞ = ∞, lim x ∞ = 0
21. f(x) =
: d: (-∞, ∞), r: (0, 1000), bounded, no max/min, continuous, increasing: (-∞, ∞), neither,
VA: none, HA: y = 0 and y = 1000, lim x -∞ = 0, lim x ∞ = 1000
22. find the y-intercepts, HA’s and sketch the following: f(x) =
intercept: (0, .57)
: HA: y = 0, y = 8, y-
Part 2: calculator part
Use a calculator to evaluate:
23. log 5: .699
24. log -3: undefined
25. log 3 + 5: 5.477
26. ln 8: 2.079
27. log3 7: 1.77
28. log2 9: 3.167
29. The population in a town in 2000 was 24,750. If the population is increasing at a rate of 2.5% per year
find the following:
a. Write the exponential function: A = 24,750(1 + .025)t, A = 24750(1.025)t
b. What = would the population be in 2004?: A = 24750(1.025)4, A = 27319
c. In what year will the population reach 40,000: in 19 years so in the year 2019
30. 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 = 6.5(
b. How much is left after 10 days? : 5 days
c. When will there be less than 5 grams remaining?: 14 days
31. The population of a city in the year 2000, can be modeled as the following: f(x) = 125,000(.975)t
a. What is the population in the year 2000?: 125,000
b. Is the population increasing or decreasing?: decreasing (because .975 is less than 1)
c. At what rate is the population increasing or decreasing?: 1- .975 = .025, 2.5%
32. The number of students infected with the flu at a high school after t days is modeled by the function
P(t) =
.
a. What has the initial number of infected students?: 19
b. What is the maximum number of students that can be infected? Why?: 800, because that is the
numerator is the limit
c. The school will close when 350 of the 800 student body are infected. When will the school close?:
after 18 days
Solve the following equations. Round to the nearest thousandth
33. 32x = 27: 32x = 33, 2x = 3, x = 1.5
34. 2(1/4)4x = 32: (1/4)4x = 16, (4)-1(4x) = 42, -4x = 2, x = -1/2
35. 5x – 4 = 10: 5x = 14, log5 5x = log5 14, x = log5 14, x = 1.640 (use change of base formula)
36. 6-2x = 18: log 6 6-2x = log6 18, -2x = log6 18, -2x = 1.613…, x = -.807 (remember don’t round until you get
your final answer)
37. log2 32 = x: 2x = 32, 2x = 25, x = 5
38. log4 x-4 = 4:
, x-4 = 256, x = 260
39. 4 + 3 ln (x – 3) = 8: 3 ln (x – 3) = 4, ln (x- 3) = 4/3, eln x – 3 =
, x – 3 = 3.79…, x = 6.794
40. -2 ln 4x = 5: ln 4x = -5/2, eln 4x =
, 4x = .082…, x = .021
-2x
-2x
41. e = 10: ln e = ln 10, -2x = 2.302…, x = -1.151
42. 4 - 2ex = 3: -2ex = -1, ex = ½, ln ex = ln ½, x = -.693
43. You deposit $400 into an account that pays 3.5% simple interest. After 5 years, how much will you
have?
A = 400(1 + .035)5 = $475.07
44. You deposit $4500 into an account that pays 2.75% interest compounded quarterly. After 3 years, how
much will you have?
A = 4500(1 +
= $4885.61
45. 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) = $1211.67
46. What is the interest rate compounded monthly is required for $500 to reach $800 in 5 years?
800 = 500(1 +
, 1.6 = (1 +
, 1.02… = 1 + r/4, .023… = r/4, .095… = r, 9.51%
47. Find the APY for an investment with an interest rate of 8.25% compounded monthly.
(remember P doesn’t matter since they will cancel out)
P(1 + x) = P (1 +
, 1 + x = 1.085…, x = .085…, 8.57%
48. Find the future value of a retirement account in which Zachary makes payments of $150 in which
interest is credited quarterly with an interest rate of 6.5% after 25 years.
FV = 150 (
), FV = $19,837.62
49. Find the present value of a loan in which payments of $247.60 are made quarterly with an interest rate
of 3.25% for 6 years.
PV = 247.50(
, $5376.97
50. How much are the monthly payments on a loan in which $14,500 is borrowed at a rate of 2.25% for 3
years.
14500(.025/12) = R (1 – (1 +
) =$418.49
51. You contribute $50 per month into an account that earns 4.5% annual interest. After 30 years, how
much will you have?
FV = 50(
) = $$37,969.31