Exam

Name___________________________________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Solve the problem.

1) 1)An initial investment of $14,000 is invested for 9 years in an account that earns 10% interest,compounded semiannually. Find the amount of money in the account at the end of the period.

A) $33,692.67 B) $33,011.27 C) $32,088.26 D) $19,692.67

2) 2)An initial investment of $12,000 is invested for 5 years in an account that earns 7% interest,compounded quarterly. Find the amount of money in the account at the end of the period.

A) $16,685.34 B) $16,830.62 C) $16,977.34 D) $4977.34

3) 3)Suppose that $2200 is invested at 4% interest, compounded quarterly. Find the function for theamount of money after t years.

A) A(t) = 2200(1.04)4t B) A(t) = 2200(1.01)4t

C) A(t) = 2200(1.01)t D) A(t) = 2200(1.04)t

4) 4)The number of bacteria growing in an incubation culture increases with time according toB = 2000(5)x, where x is time in days. Find the number of bacteria when x = 0 and x = 2.

A) 2000, 6,250,000 B) 2000, 50,000 C) 10,000, 50,000 D) 2000, 20,000

5) 5)Suppose the amount of a radioactive element remaining in a sample of 100 milligrams after x yearscan be described by A(x) = 100e-0.01731x. How much is remaining after 176 years? Round theanswer to the nearest hundredth of a milligram.

A) 304.66 milligrams B) 0.05 milligrams

C) 4.75 milligrams D) 2104.28 milligrams

6) 6)The value of a stock is given by the function

V(t) = 67(1 - e-1.3t) + 30,

where V is the value of the stock after time t, in months. When will the value of the stock be $93?

A) 0.6 months B) 1.5 months C) 2.2 months D) 2.8 months

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7) 7)A company begins a radio advertising campaign in Chicago to market a new soft drink. Thepercentage of the target market that buys a soft drink is estimated by the function

f(t) = 100(1 - e-0.01t),

where t is the number of days of the campaign.

After how long will 90% of the target market have bought the soft drink?

A) 3 days B) 90 days C) 231 days D) 230 days

Use a graphing calculator. Solve graphically.

8) 8)10-x < 1

A) (0, ) B) [0, ) C) (- , 0] D) (- , 0)

9) 9)3.7x - 4.8x - 2 = 200

A) 4 B) 2.154 C) 4.154 D) No solution

Solve the problem.

10) 10)In September 1998 the population of the country of West Goma in millions was modeled byf(x) = 17.1e0.0017x. At the same time the population of East Goma in millions was modeled by g(x) = 13.6e0.0140x. In both formulas x is the year, where x = 0 corresponds to September 1998.Assuming these trends continue, estimate the year when the population of West Goma will equalthe population of East Goma.

A) 19 B) 1979 C) 2013 D) 2017

Use a graphing calculator. Solve graphically.

11) 11)e2x < x4

A) (- , -0.703] B) (- , -0.703) C) [-0.703, ] D) (-0.703, ]

Find the value of the expression.

12) 12)log22

1

A) 10 B) 0 C) 1 D) 22

13) log 51

625 13)

A) 20 B) 5 C) -4 D) 1/4

14) 14)log3 3

A) 12 B) - 2 C) -

12 D) 2

2

15) 15)ln e-6

A) -16 B) -6 C) 1 D) e ln -6

Convert to a logarithmic equation.

16) 16)32 = 9

A) 2 = log 3 9 B) 3 = log 2 9 C) 9 = log 4 2 D) 2 = log 16 3

17) 17)e-7 = 0.0009119

A) e = log-7 0.0009119 B) 0.0009119 = log e -7

C) 0.0009119 = log-7 e D) -7 = log e 0.0009119

18) 18)100.699 = 5

A) 0.699 = log 10 5 B) 10 = log 9 0.699 C) 5 = log 10 0.699 D) 0.699 = log 9 10

19) 3-2 =19 19)

A) 19

= log 3 -2 B) -2 = log 1/9 3 C) 3 = log 219 D) -2 = log 3

19

Convert to an exponential equation.

20) 20)log7 1 = 0

A) 07 = 1 B) 10 = 7 C) 71 = 0 D) 70 = 1

21) 21)ln 44 = 3.7842

A) e44 = 3.7842 B) e3.7842 = 1 C) e3.7842 = ln 44 D) e3.7842 = 44

22) 22)log w Q = 20

A) Q20 = w B) Qw = 20 C) w20 = Q D) 20w = Q

Find the logarithm using the change-of-base formula.

23) 23)log6 5

A) 1.1133 B) 1.7965 C) -0.8982 D) 0.8982

24) 24)log7 58.28

A) 8.3257 B) 0.4787 C) 1.7655 D) 2.0891

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25) 25)log23 76.03

A) 3.3057 B) 1.8810 C) 1.3813 D) 0.7239

Find the domain and the vertical asymptote of the function.

26) 26)f(x) = log (x - 5)

A) Domain (5, ); vertical asymptote: x = 5

B) Domain (1, ); vertical asymptote: x = 1

C) Domain (0, ); vertical asymptote: x = 0

D) Domain : (-5, ); vertical asymptote: x = -5

27) 27)f(x) = 4 + log6x

A) Domain: [0, ); vertical asymptote: x = 0 B) Domain: (0, ); vertical asymptote: x = 0

C) Domain: (4, ); vertical asymptote: x = 4 D) Domain: (0, ); vertical asymptote: none

28) 28)f(x) = ln x - 2

A) Domain: (- 2, ); vertical asymptote: x = - 2

B) Domain: (- , ); vertical asymptote: none

C) Domain: [0, ); vertical asymptote: x = 0

D) Domain: (0, ); vertical asymptote: x = 0

Solve.

29) 29)An earthquake was recorded with an intensity which was 50,119 times more powerful than areference level earthquake, or 50,119 · Io. What is the magnitude of this earthquake on the Richterscale (rounded to the nearest tenth)? The magnitude on the Richter scale of an earthquake ofintensity I is log10(I/Io).

A) 0.5 B) 4.7 C) 3.7 D) 10.8

30) 30)The loudness, D, in decibels of a sound of intensity S is given byD = 10 log (S/So), where S is measured in watt/m2 and So = 10-12 watt/m2. What is the

decibel level of a noise whose intensity is 9.02 x 10-5 watt/m2? (Round to the nearest wholenumber.)

A) 8 decibels B) 183 decibels C) 70 decibels D) 80 decibels

31) 31)In chemistry, the pH of a substance is defined by pH = -log [H+], where [H+] is the hydrogen ionconcentration in moles per liter. Find the pH of a sample of lake water whose hydrogen ionconcentration, [H+], is 3.05 x 10-9 moles per liter. (Round to the nearest tenth.)

A) 6.4 B) 8.5 C) 7.3 D) 10.1

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32) 32)Find the hydrogen ion concentration of a solution whose pH is 7.4. Use the formula pH = -log [H+].

A) 3.98 x 10-8 B) -0.8692317 C) 0.86923172 D) 2.51 x 107

Express in terms of sums and differences of logarithms.

33) 33)logax2yz2

A) loga2x + logay + loga2z B) 2 logax · logay · 2logaz

C) 2 logax + logay + 2logaz D) (logax)2 + logay + (logaz) 2

34) log 3x5

y634)

A) log 3 + 5 log x - 6 log y B) log (3 + x5 - y6)

C) log 3 + (log x)5 - (log y)6 D) log 3 · 5 log x - 6 log y

35) logbm7p2

n4b635)

A) logb (m7p2 - n4b6) B) 7logbm · 2logbp ÷ 4logbn - 6

C) 7logbm + 2logbp - 4logbn - 6 D) 7logbm + 2logbp - 4logbn + 6

36) logbx6y2

z7 36)

A) 3logbx + 2logby - 7logbz B) 3logbx - logby +72

logbz

C) 3logbx · logby ÷ 72

logbz D) 3logbx + logby -72

logbz

Express as a single logarithm and, if possible, simplify.

37) 37)loga0.1 + loga 100

A) loga1

1000 B) loga 10

C) 0.1 loga 100 D) loga 0.1 · loga 100

38) 12

log s + 2log m 38)

A) log (m2 + s ) B) log sm22

C) log s2

+ 2m D) log m2 s

5

39) 12

logax + 5 loga y - 3 loga x 39)

A) loga x y5 B) logay5

x5/2C) loga x3 y5 D) loga x5 y5

40) 40)ln (x2 - 64) - ln (x + 8)

A) ln (x2 - 8) B) ln (x + 8) C) ln (x - 8) D) ln (x - 64)

41) 41)logb x8 - 2logb x

A) logb x7 B) logb x9 C)logb x8

logb xD) logb (x8 - x)

42) log a4x

- log a 4x 42)

A) log a2x B) log a

x4

C) log a4x

- 4x D) log a1x

Solve.

43) Given that loga3 = 1.099, and loga5 = 1.609, find loga35

. 43)

A) -0.51 B) 0.511 C) 2.708 D) 0.683

44) 44)Given that loga7 = 0.845, and loga3 = 0.477, find loga49.

A) 0.714 B) 1.322 C) 0.403 D) 1.69

45) 45)Given log b 6 = 1.3695 and log b 7 = 1.4873, evaluate log b 6b .

A) 1.3695 B) 2.3695 C) 2.8568 D) 1.1956 + b

Simplify.

46) 46)logaa4

A) 1 B) 4logaa C) a4 D) 4

47) 47)7log7 (4x)

A) 1 B) 74x C) 7 D) 4x

48) 48)log e e x - 22

A) log e 22 B) 22log e e C) log x - 22 D) x - 22

6

49) 49)log b b3 .

A) 12 B) 3 C) 2 D) 3

2

Solve the exponential equation.

50) 50)2(8 - 2x) = 4

A) 2 B) -3 C) 4 D) 3

51) 51)97x = 81

A) 512 B) 72 C) 2

7 D) 81

52) 52)72x = 70Give your answer in exact form.

A) log 70log 72 B) log 72

log 70 C) log 70 D) log 72

53) 53)e-x = 32x

A) 3 B) No solution C) 0 D) 12

ln 3

54) 54)ex + e-x = 4

A) 0.9115, -1.7224 B) 1.4436, -1.4436 C) 1.317, -1.317 D) 1.297, -1.0739

55) ex + e-x

ex - e-x= 2 55)

A) 0.5616 B) 0.5493 C) 0.6931 D) 0.6727

Solve the logarithmic equation.

56) log 9 x =12 56)

A) 81 B) 512 C) 3 D) 0.00195312

57) 57)ln x = 3Give your answer in exact form.

A) e3 B) 1000 C) ln 3 D) 3e

58) 58)log (x + 9) = 1 - log x

A) 1 B) -1, 10 C) -1 D) -10, 1

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59) 59)log4 (x - 7) + log4 (x - 7) = 1

A) -9, 9 B) 50 C) - 50, 50 D) 9

60) 60)log ( 3 + x) - log (x - 4 ) = log 4

A) B) 193 C) -

193 D) 5

2

61) 61)log (x + 10) - log (x + 4) = log x

A) 2 B) 6 C) 2, -5 D) No solution

Find approximate solutions of the equation.

62) 62)xe2x - 1 = 1

A) 0.347 B) 0.601 C) 0.853 D) No solution

63) 63)log4(2x + 5) - log4(x - 2) = 1

A) 2.408 B) 3.125 C) 6.5 D) No solution

Use a graphing calculator to find the approximate point(s) of intersection of the pair of equations.

64) 64)y = ln 2x; y = 2x - 8

A) (0.017, -7.968), (5.077, 2.077) B) (0.00017, -7.9997), (5.168, 2.336)

C) (0.00017, -7.9997), (5.077, 2.336) D) (5.168, 2.336)

65) 65)y = 2.3 ln (x + 10.7); y = e-0.007x2

A) (-9.8111, 0.9096) B) (-9.4375, .5360)

C) (-9.8111, 0.1624) D) (-9.0639, 0.9096)

Solve.

66) 66)When interest is compounded continuously, the balance in an account after t years is given byP t = P0ekt, where P0 is the initial investment and k is the interest rate. Suppose that P0 is investedin a savings account where interest is compounded continuously at 9% per year. Express P t interms of P0 and 0.09.

A) P t = P0e0.09t B) P t = P 0.09t0 C) P t = P0e0.09 D) P t = P0e9t

67) 67)Under ideal conditions, a population of rabbits has an exponential growth rate of 11.8% per day.Consider an initial population of 900 rabbits. Find the exponential growth function.

A) P(t) = 90e0.118t B) P(t) = 100e11.8t

C) P(t) = 100e1.18t D) P(t) = 900e0.118t

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68) 68)In 1996, an outbreak of a disease infected 20 people in a large community. By 1997, the number ofthose infected had grown to 36. Find an exponential growth function that fits the data. (Rounddecimals to three places.)

A) N t = 20e5.88t, where t is the number of years after 1997.

B) N t = 20e5.88t, where t is the number of years after 1996.

C) N t = 20e0.588t, where t is the number of years after 1996.

D) N t = 20e0.588t, where t is the number of years after 1997.

69) 69)How long will it take for the population of a certain country to double if its annual growth rate is7.5%? Round to the nearest year.

A) 1 yr B) 27 yr C) 4 yr D) 9 yr

70) 70)There are currently 50 million cars in a certain country, decreasing by 4.9% annually. How manyyears will it take for this country to have 27 million cars? Round to the nearest year.

A) 2 years B) 64 years C) 13 years D) 5 years

71) 71)Susan purchased a painting in the year 2000 for $3000. Assuming an exponential rate of inflation of2.4% per year, how much will the painting be worth 7 years later?

A) $4891.02 B) $9176.03 C) $7812.90 D) $3548.81

72) 72)How long will it take for $8000 to grow to $45,300 at an interest rate of 11% if the interest iscompounded continuously? Round the number of years to the nearest hundredth.

A) 1.58 yr B) 1576.24 yr C) 0.16 yr D) 15.76 yr

73) 73)Kimberly invested $3000 in her savings account for 8 years. When she withdrew it, she had$3693.64. Interest was compounded continuously. What was the interest rate on the account?

A) 2.6% B) 2.75% C) 2.5% D) 2.7%

74) 74)In a town whose population is 1500, a disease creates an epidemic. The number N of peopleinfected t days after the disease has begun is given by the function

N(t) =1500

1 + 20.2 · e-0.6t

Find the number infected after 8 days.

A) 1286 B) 1284 C) 1289 D) 1288

Solve the problem.

75) 75)A sample of 250 grams of radioactive substance decays according to the function A(t) = 250e-0.032t,where t is the time in years. How much of the substance will be left in the sample after 30 years?Round to the nearest whole gram.

A) 0 grams B) 1 gram C) 96 grams D) 524 grams

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76) 76)An artifact is discovered at a certain site. If it has 70 % of the carbon-14 it originally contained,what is the approximate age of the artifact? (carbon-14 decays at the rate of 0.0125% annually.)

A) 5600 years B) 2853 years C) 2400 years D) 1239 years

77) 77)The number of acres in a landfill is given by the function B = 3000e-0.04t, where t is measured inyears. How many acres will the landfill have after 9 years? (Round to the nearest acre.)

A) 3065 B) 1310 C) 2093 D) 1331

78) 78)Newton's Law of Cooling states that if a body with temperature T1 is placed in surroundings withtemperature T0 different from T1, then the body will either cool or warm to temperature T(t) after

time t, in minutes, where T(t) = T0 + (T1 - T0)e-kt.A cup of coffee with temperature 104°F is placed in a freezer with temperature 0°F. After 8minutes, the temperature of the coffee is 53.1°F. What will its temperature be 14 minutes after it isplaced in the freezer? Round your answer to the nearest degree.

A) 27°F B) 30°F C) 25°F D) 32°F

Convert to an exponential equation.

79) 79)log2 16 = 4

A) 164 = 2 B) 216 = 4 C) 42 = 16 D) 24 = 16

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