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

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1 Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Solve the problem. 1) An initial investment of $14,000 is invested for 9 years in an account that earns 10% interest, 1) compounded semiannually. Find the amount of money in the account at the end of the period. A) $33, B) $33, C) $32, D) $19, ) An initial investment of $12,000 is invested for 5 years in an account that earns 7% interest, 2) compounded quarterly. Find the amount of money in the account at the end of the period. A) $16, B) $16, C) $16, D) $ ) Suppose that $2200 is invested at 4% interest, compounded quarterly. Find the function for the 3) amount 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) The number of bacteria growing in an incubation culture increases with time according to 4) B = 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) Suppose the amount of a radioactive element remaining in a sample of 100 milligrams after x years 5) can be described by A(x) = 100e x. How much is remaining after 176 years? Round the answer to the nearest hundredth of a milligram. A) milligrams B) 0.05 milligrams C) 4.75 milligrams D) milligrams 6) The value of a stock is given by the function 6) 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 1

2 7) A company begins a radio advertising campaign in Chicago to market a new soft drink. The 7) percentage 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) 10-x < 1 8) A) (0, ) B) [0, ) C) (-, 0] D) (-, 0) 9) 3.7x - 4.8x - 2 = 200 9) A) 4 B) C) D) No solution Solve the problem. 10) In September 1998 the population of the country of West Goma in millions was modeled by 10) f(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 Assuming these trends continue, estimate the year when the population of West Goma will equal the population of East Goma. A) 19 B) 1979 C) 2013 D) 2017 Use a graphing calculator. Solve graphically. 11) e2x < x4 11) A) (-, ] B) (-, ) C) [-0.703, ] D) (-0.703, ] Find the value of the expression. 12) log 1 12) 22 A) 10 B) 0 C) 1 D) 22 13) log ) A) 20 B) 5 C) -4 D) 1/4 14) log3 3 14) A) 1 2 B) - 2 C) D) 2 2

3 15) ln e-6 15) A) B) -6 C) 1 D) e ln -6 Convert to a logarithmic equation. 16) 32 = 9 16) A) 2 = log 3 9 B) 3 = log 2 9 C) 9 = log 4 2 D) 2 = log ) e-7 = ) A) e = log B) = log e -7 C) = log -7 e D) -7 = log e ) = 5 18) A) = log 10 5 B) 10 = log C) 5 = log D) = log ) 3-2 = ) A) 1 9 = log 3-2 B) -2 = log 1/9 3 C) 3 = log D) -2 = log Convert to an exponential equation. 20) log 7 1 = 0 20) A) 07 = 1 B) 10 = 7 C) 71 = 0 D) 70 = 1 21) ln 44 = ) A) e44 = B) e = 1 C) e = ln 44 D) e = 44 22) log Q = 20 22) w A) Q20 = w B) Qw = 20 C) w20 = Q D) 20w = Q Find the logarithm using the change-of-base formula. 23) log6 5 23) A) B) C) D) ) log ) A) B) C) D)

4 25) log ) A) B) C) D) Find the domain and the vertical asymptote of the function. 26) f(x) = log (x - 5) 26) 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) f(x) = 4 + log 6 x 27) 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) f(x) = ln x ) 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) An earthquake was recorded with an intensity which was 50,119 times more powerful than a 29) reference level earthquake, or 50,119 Io. What is the magnitude of this earthquake on the Richter scale (rounded to the nearest tenth)? The magnitude on the Richter scale of an earthquake of intensity I is log10(i/io). A) 0.5 B) 4.7 C) 3.7 D) ) The loudness, D, in decibels of a sound of intensity S is given by 30) D = 10 log (S/So), where S is measured in watt/m2 and So = watt/m2. What is the decibel level of a noise whose intensity is 9.02 x 10-5 watt/m2? (Round to the nearest whole number.) A) 8 decibels B) 183 decibels C) 70 decibels D) 80 decibels 31) In chemistry, the ph of a substance is defined by ph = -log [H+], where [H+] is the hydrogen ion 31) concentration in moles per liter. Find the ph of a sample of lake water whose hydrogen ion concentration, [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)

5 32) Find the hydrogen ion concentration of a solution whose ph is 7.4. Use the formula ph = -log [H+]. 32) A) 3.98 x 10-8 B) C) D) 2.51 x 107 Express in terms of sums and differences of logarithms. 33) logax2yz2 33) A) loga2x + logay + loga2z B) 2 logax logay 2logaz C) 2 logax + logay + 2logaz D) (logax) 2 + logay + (logaz) 2 34) log 3x 5 y6 34) A) log 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) logb m 7p2 n4b6 35) A) logb (m7p2 - n4b6) B) 7logbm 2logbp 4logbn - 6 C) 7logbm + 2logbp - 4logbn - 6 D) 7logbm + 2logbp - 4logbn ) logb x 6y2 z7 36) A) 3logbx + 2logby - 7logbz B) 3logbx - logby log bz C) 3logbx logby 7 2 log bz D) 3logbx + logby log bz Express as a single logarithm and, if possible, simplify. 37) loga0.1 + loga ) A) loga B) loga 10 C) 0.1 loga 100 D) loga 0.1 loga ) 1 log s + 2log m 38) 2 A) log (m2 + s ) B) log sm 2 2 C) log s 2 + 2m D) log m 2 s 5

6 39) 1 2 log ax + 5 loga y - 3 loga x 39) A) loga x y5 B) loga y 5 x5/2 C) loga x3 y5 D) loga x5 y5 40) ln (x2-64) - ln (x + 8) 40) A) ln (x2-8) B) ln (x + 8) C) ln (x - 8) D) ln (x - 64) 41) logb x8-2logb x 41) A) logb x7 B) logb x9 C) log b x8 D) logb x logb (x8 - x) 42) log a 4 x - log a 4x 42) A) log a 2 x C) log a B) log a 4 x - 4x D) log a 1 x x 4 Solve. 43) Given that loga3 = 1.099, and loga5 = 1.609, find loga ) A) B) C) D) ) Given that loga7 = 0.845, and loga3 = 0.477, find loga49. 44) A) B) C) D) ) Given log b 6 = and log b 7 = , evaluate log b 6b. 45) A) B) C) D) b Simplify. 46) logaa4 46) A) 1 B) 4logaa C) a4 D) 4 47) 7log7 (4x) 47) A) 1 B) 74x C) 7 D) 4x 48) log e e x ) A) log e 22 B) 22log e e C) log x - 22 D) x

7 49) log b b3. 49) A) 1 2 B) 3 C) 2 D) 3 2 Solve the exponential equation. 50) 2(8-2x) = 4 50) A) 2 B) -3 C) 4 D) 3 51) 97x = 81 51) A) 512 B) 7 2 C) 2 7 D) 81 52) 72x = 70 52) Give your answer in exact form. A) log 70 log 72 B) log 72 log 70 C) log 70 D) log 72 53) e-x = 32x 53) A) 3 B) No solution C) 0 D) 1 2 ln 3 54) ex + e-x = 4 54) A) , B) , C) 1.317, D) 1.297, ) e x + e-x ex - e-x = 2 55) A) B) C) D) Solve the logarithmic equation. 56) log 9 x = ) A) 81 B) 512 C) 3 D) ) ln x = 3 57) Give your answer in exact form. A) e3 B) 1000 C) ln 3 D) 3e 58) log (x + 9) = 1 - log x 58) A) 1 B) -1, 10 C) -1 D) -10, 1 7

8 59) log 4 (x - 7) + log 4 (x - 7) = 1 59) A) -9, 9 B) 50 C) - 50, 50 D) 9 60) log ( 3 + x) - log (x - 4 ) = log 4 60) A) B) 19 3 C) D) ) log (x + 10) - log (x + 4) = log x 61) A) 2 B) 6 C) 2, -5 D) No solution Find approximate solutions of the equation. 62) xe2x - 1 = 1 62) A) B) C) D) No solution 63) log 4 (2x + 5) - log 4 (x - 2) = 1 63) A) B) C) 6.5 D) No solution Use a graphing calculator to find the approximate point(s) of intersection of the pair of equations. 64) y = ln 2x; y = 2x ) A) (0.017, ), (5.077, 2.077) B) ( , ), (5.168, 2.336) C) ( , ), (5.077, 2.336) D) (5.168, 2.336) 65) y = 2.3 ln (x ); y = e-0.007x2 65) A) ( , ) B) ( ,.5360) C) ( , ) D) ( , ) Solve. 66) When interest is compounded continuously, the balance in an account after t years is given by 66) P t = P0ekt, where P0 is the initial investment and k is the interest rate. Suppose that P0 is invested in a savings account where interest is compounded continuously at 9% per year. Express P t in terms of P0 and A) P t = P0e0.09t B) P t = P 0.09t 0 C) P t = P0e0.09 D) P t = P0e9t 67) Under ideal conditions, a population of rabbits has an exponential growth rate of 11.8% per day. 67) 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 8

9 68) In 1996, an outbreak of a disease infected 20 people in a large community. By 1997, the number of 68) those infected had grown to 36. Find an exponential growth function that fits the data. (Round decimals to three places.) A) N t = 20e5.88t, where t is the number of years after B) N t = 20e5.88t, where t is the number of years after C) N t = 20e0.588t, where t is the number of years after D) N t = 20e0.588t, where t is the number of years after ) How long will it take for the population of a certain country to double if its annual growth rate is 69) 7.5%? Round to the nearest year. A) 1 yr B) 27 yr C) 4 yr D) 9 yr 70) There are currently 50 million cars in a certain country, decreasing by 4.9% annually. How many 70) years 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) Susan purchased a painting in the year 2000 for $3000. Assuming an exponential rate of inflation of 71) 2.4% per year, how much will the painting be worth 7 years later? A) $ B) $ C) $ D) $ ) How long will it take for $8000 to grow to $45,300 at an interest rate of 11% if the interest is 72) compounded continuously? Round the number of years to the nearest hundredth. A) 1.58 yr B) yr C) 0.16 yr D) yr 73) Kimberly invested $3000 in her savings account for 8 years. When she withdrew it, she had 73) $ 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) In a town whose population is 1500, a disease creates an epidemic. The number N of people 74) infected t days after the disease has begun is given by the function N(t) = e-0.6t Find the number infected after 8 days. A) 1286 B) 1284 C) 1289 D) 1288 Solve the problem. 75) A sample of 250 grams of radioactive substance decays according to the function A(t) = 250e-0.032t, 75) 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 9

10 76) An artifact is discovered at a certain site. If it has 70 % of the carbon-14 it originally contained, 76) what is the approximate age of the artifact? (carbon-14 decays at the rate of % annually.) A) 5600 years B) 2853 years C) 2400 years D) 1239 years 77) The number of acres in a landfill is given by the function B = 3000e-0.04t, where t is measured in 77) years. How many acres will the landfill have after 9 years? (Round to the nearest acre.) A) 3065 B) 1310 C) 2093 D) ) Newton's Law of Cooling states that if a body with temperature T1 is placed in surroundings with 78) temperature 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 8 minutes, the temperature of the coffee is 53.1 F. What will its temperature be 14 minutes after it is placed 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) log 2 16 = 4 79) A) 164 = 2 B) 216 = 4 C) 42 = 16 D) 24 = 16 10