# Math Problems

Only available on StudyMode
• Published : February 7, 2012

Text Preview
1. Solve
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%?