# Mathematics and Nearest Cent

a. e^(.05t) = 1600

Answer: 147.56

Show your work in this space:

ln〖e^0.05t=ln1600 〗

0.05t= ln1600

t= ln1600/0.05=147.56

b. ln(4x) =3

Answer: x= e^3/4

Show your work in this space:

e^ln4x =e^3

4x=e^3

x= e^3/4

c. log_2〖(8-6x)〗= 5

Answer: x=-4

Show your work in this space:

8-6x=2^5

-6x=32-8

-6x=24

x=-4

d. 4 + 5e^(‐x) = 0

Answer: No solution.

Show your work in this space:

5e^(-x)=-4

e^(-x)=-4/5

ln〖e^(-x) 〗=ln(-4/5)=undefined

Since the right side is undefined, there is no solution.

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

Description of transformation: The original function (f(x) = log x) is shifted to the left 5 units horizontally. Equation(s) for the Vertical Asymptote(s): x = -5

x+5=0

x=-5

X-intercept in (x, y) form: (-4,0)

Set g(x) = 0:

log(x+5)=0

x+5=〖10〗^0

x+5=1

x=-4

b) g(x)=log〖(-x)〗

Description of transformation: reflect about the y-axis

Equation(s) for the Vertical Asymptote(s): x = 0

-x=0

x=0

X-intercept in (x, y) form: (-1,0)

Set g(x) = 0:

log(-x)=0

-x=〖10〗^0

x=-1

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

Answer: 68

Show your work in this space:

If t = 0 then

S(0)=68-20log〖(0+1)〗=68-20(0)=68-0=68

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

Answer: S(4) = 54.02; S(24) = 40.04

Show your work in this space:

If t = 4...

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