Math Problems
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%?
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
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