Question

the formula ( a = pe^{rt} ) describes the accumulated value, ( a ), of a sum of money, ( p ), the principal, after ( t ) years at annual percentage rate ( r ) (in decimal form) compounded continuously, or a savings account subject to continuous compounding.

amount invested	annual interest rate	accumulated amount	time ( t ) in years

(do not round until the final answer. then round to one decimal place as needed.)

Explanation:

Step1: Recall the continuous compounding formula

The formula for continuous compounding is \( A = Pe^{rt} \), where \( A \) is the accumulated amount, \( P \) is the principal (amount invested), \( r \) is the annual interest rate (in decimal), and \( t \) is the time in years. We want to find the time \( t \) when \( A = 2P \) (since we want to double the amount invested).

Step2: Substitute \( A = 2P \) and \( r = 0.13 \) into the formula

Substituting \( A = 2P \) and \( r = 0.13 \) into \( A = Pe^{rt} \), we get:

$$ 2P = Pe^{0.13t} $$

We can divide both sides by \( P \) (since \( P
eq 0 \)):

$$ 2 = e^{0.13t} $$

Step3: Take the natural logarithm of both sides

To solve for \( t \), we take the natural logarithm (ln) of both sides:

$$ \ln(2) = \ln(e^{0.13t}) $$

Using the property of logarithms \( \ln(e^x) = x \), the right side simplifies to \( 0.13t \):

$$ \ln(2) = 0.13t $$

Step4: Solve for \( t \)

Now, we solve for \( t \) by dividing both sides by \( 0.13 \):

$$ t = \frac{\ln(2)}{0.13} $$

Calculating \( \ln(2) \approx 0.6931 \), we have:

$$ t \approx \frac{0.6931}{0.13} \approx 5.3315 $$

Rounding to one decimal place, \( t \approx 5.3 \)

Answer:

\( t \approx 5.3 \) years