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1) a. e.05t = 1600 Answer: t = 147.556 Work: e^(05t) = 1600 or, 0.05t = ln(1600) or, 0.05t = 7.3778 or, t = 7.3778/0.5 or, t = 147.556 b. ln(4x) =3 Answer: x = 5.0214 Work: ln(4x) =3 or, 4x = e^3 or, x = e^3/4 or, x = 5.0214 c. log2(8 – 6x) = 5 Answer: x = -4 Work: log2(8-6x) = 5 or, 8-6x = 2^5 or, 8-6x = 32 or, 6x = 8-32 or, x = -24/6 or, x = -4 d. 4 + 5e-x = 0 Answer: No solution or, 5e^-x = -4 or, e^-x = -4/5 Here left side is positive for all values of x, while right is negative. So, no solution exists. 2) Describe the transformations on the following graph of)log()(xxf=. State the placement of the vertical asymptote and x-intercept after the transformation. For example, vertical shift up 2 or reflected about the x-axis are descriptions. XY-10-9-8-7-6-5-4-3-2-112345678910-10-9-8-7-6-5-4-3-2-1123456789100 a) g(x) = log( x + 5) Description of transformation: Left shift 5 and take log Equation(s) for the Vertical Asymptote(s): x = -5 x-intercept in (x, y) form: (-4, 0) b) ) g(x)=log(-x) Description of transformation: Reflect about y-axis and take log Equation(s) for the Vertical Asymptote(s): x = 0 x-intercept in (x, y) form: (-1, 0) 3. Students in an English class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. The average score S(t), in percent, after t months was found to be given by S(t) = 68 - 20 log (t + 1), t = 0. What was the average score when they initially took the test, t = 0? Round your answer to a whole percent, if necessary. a) Answer: 68% Show your work in this space: S(0) = 68-20*log(0+1) = 68-20*0 = 68 What was the average score after 4 months? after 24 months? Round your answers to two decimal places. b) Answer: After 4 months: 54.02%, After 24 months: 40.04% Show your work in this space: S(4) = 68-20*log(4+1) = 54.02 S(24) = 68-20*log(24+1) = 40.04 After what time t was the average score 50%? Round your answers to two decimal places. c) Answer: t = 6.94 Show your work in this space: 50 = 68 - 20 log (t + 1) Or, 20log(t+1) = 68-50 Or, log(t+1) = 18/20 Or, t+1 = 10^(18/20) = 7.9433 Or, t = 7.9433-1 = 6.94 4) The formula for calculating the amount of money returned for an initial deposit into a bank account or CD (certificate of deposit) is given by ntnrPA??????+=1 A is the amount of the return. P is the principal amount initially deposited. r is the annual interest rate (expressed as a decimal). n is the number of compound periods in one year. t is the number of years. Carry all calculations to six decimals on each intermediate step, then round the final answer to the nearest cent. Suppose you deposit $2,000 for 5 years at a rate of 8%. a) Calculate the return (A) if the bank compounds annually (n = 1). Round your answer to the nearest cent. Answer: A = $2,938.66 Show work in this space. Use ^ to indicate the power or use the Equation Editor in MS Word. A = $2000*(1+0.08/1)^(1*5) = $2,938.66 b) Calculate the return (A) if the bank compounds quarterly (n = 4). Round your answer to the nearest cent. Answer: A = $2,971.89 Show work in this space: A = $2000*(1+0.08/4)^(4*5)= $2,971.89 c) If a bank compounds continuously, then the formula used is rtPeA= where e is a constant and equals approximately 2.7183. Calculate A with continuous compounding. Round your answer to the nearest cent. Answer: A = $2,983.66 Show work in this space: A = $2000*(2.7183)^(0.08*5) = $2,983.66

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posted: | 12/4/2011 |

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