Question

Michelle deposited $1,600 into an account that is compounded continuously with an annual interest rate of 8%. After t years, there was $3,200 in the account. Based on the given information, if the formula for continuously compounded interest is A = P e r t , where P represents the initial amount of money, r represents the annual interest rate (as a decimal), t represents the number of years, and A represents the amount of money after t years, how many years had the money been in the account? Express your answer as a logarithm.

Answer 1

Answer:

As a logarithm, the money had been in the account for $t = \frac{\ln{2}}{0.08}$ years.

Step-by-step explanation:

The amount of money in the account after t years is given by:

$A(t) = Pe^{rt}$

In which P is the initial amount deposited and r is the interest rate, as a decimal.

Michelle deposited $1,600 into an account that is compounded continuously with an annual interest rate of 8%.

This means that $P = 1600, r = 0.08$

After t years, there was $3,200 in the account. How many years had the money been in the account?

This is t for which $P(t) = 3200$. So

$A(t) = Pe^{rt}$

$3200 = 1600e^{0.08t}$

$e^{0.08t} = \frac{3200}{1600}$

$e^{0.08t} = 2$

$\ln{e^{0.08t}} = \ln{2}$

$0.08t = \ln{2}$

$t = \frac{\ln{2}}{0.08}$

$t = 8.67$

As a logarithm, the money had been in the account for $t = \frac{\ln{2}}{0.08}$ years.