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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Continuous schedule from Friday , November 1st. 9am until Saturday , November 2nd., 23:59pm.
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...Semi-Detailed Lesson Plan in Mathematics (Transformations)
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Subjects: Mathematics, Geometry, Transformations
I. Objectives:
A. To recognize Euclidean transformations.
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