# Math Problems

**Pages:**3 (527 words)

**Published:**February 7, 2012

a. e^.05t = 1600

0.05t = ln(1600)

0.05t = 7.378

t = 7.378/.05

t = 147.56

b. ln(4x)=3

4x = e^3

x = e^3/4

x = 5.02

c. log2(8 – 6x) = 5

8-6x = 2^5

8-6x = 32

6x = 8-32

x = -24/6

x = -4

d. 4 + 5e-x = 0

5e^(-x) = -4

e^(-x) = -4/5

no solution, e cannot have a negative answer

2. Describe the transformations on the following graph of f (x) log( x) . 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.

a. g(x) = log( x + 5)

horizontal left shift 5

Vertical asymptote x = -5

x-intercept: (-4, 0)

b. g(x)=log(-x)

over the x-axis

vertical asymptote x=0

no x-intercept

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.

a. What was the average score when they initially took the test, t = 0? Round your answer to a whole percent, if necessary. S(0)=68-20xlog(0+1) =

68-20x0

= 68%

b. What was the average score after 4 months? after 24 months? Round your answers to two decimal places.

-S(4) = 68-20xlog(4+1)

68-20x0.699

68-13.98

=54.02

-S(24) = 68-20xlog(24+1) = 40.04

68-20x1.398

68-27.96

=40.04

c. After what time t was the average score 50%?

Round your answers to two decimal places.

50 = 68 - 20 log (t + 1)

20log(t+1) = 68-50

log(t+1) = 18/20

t+1 = 10^(18/20) = 7.9433

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 A=P(1+r/n)^nt 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 decimal places on...

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