Question

An initial of the $100 is now valuated at $150. The annual interest rate is 5%, compounded continuously. The equation 100e^0.05=150 represents the situation, where t is the number of years the money has been invested. About how long has the money been invested. Use calculator to round to the nearest whole number.

Answer 1

If you need answers to the whole assignment

Explanation:

1. Consider 8^x-4 = 8^10

Because the (blank a) are equal , the (blank b) must also be equal.

Answer: Bases, Exponents

The solution to the equation is 14

2.What equation is equivalent to 9^(x-3)=729?

Answer 3^x - 3 = 3^6

Solve: 9x - 3 = 729

Answer: x = 6

3. To solve 5(2^x+4)=15, first divide each side by

Answer: 5

Solve 5(2^x+4) = 15. Round to the nearest thousandth.

Answer: -2.415

4. Which of the following is the solution of 5e^2x- 4 = 11?

Answer: x=In3/2

5. Select all of the potential solution(s) of the equation 2log5x = log54.

Answer: 2,-2

What is the solution to 2log5x = log54?

Answer: 2

6. Which equation is equivalent to log5x3 - logx2 = 2?

Answer: 10^log5^3/x^2=10^2

Solve: log5x3 - logx2 = 2

Answer: 20

7. What is the solution to ln (x2 - 16) = 0?

Answer: x=+-(17)

8. Solve: ln 2x + ln 2 = 0

Answer: ¼

Solve: e^ 2x+5 = 4

Answer: x=(In4) - 5/2

9. Consider the equation log(3x - 1) = log2(8). Explain why 3x - 1 is not equal to Describe the steps you would take to solve the equation, and state what 3x - 1 is equal to.

Answer: The bases are not the same, so you cannot set 3x - 1 equal to 8.You can evaluate the logarithm on the right side of the equation to get .You can use the definition of a logarithm to write 3x - 1 = 1000.

10. An initial investment of $100 is now valued at $150. The annual interest rate is 5%, compounded continuously. The equation 100e^0.05t = 150 represents the situation, where t is the number of years the money has been invested. About how long has the money been invested? Use your calculator and round to the nearest whole number.

Answer: 8

Answer 2

8 years approximately

The problem (I'm assuming is):

Solve
`100e^(0.05t)=150`.

I put a t in the problem where I suspected it in went.

Explanation:

`100e^(0.05t)=150`

Divide both sides by 100:

`e^(0.05t)=(150)/(100)`

`e^(0.05t)=1.5`

Rewrite in logarithm form.

`\ln(1.5)=0.05t`

Divide both sides by 0.05:

`(\ln(1.5))/(0.05)=t`

Put left hand side into calculator:

`8.109 \approx t`

`t \approx 8.109`

So about 8 years if I wrote down the equation correctly.