Question

Marshall opened a savings account 9 years ago. The account earns 1% interest, compounded continuously. If the current balance is $200.00, how much did he deposit initially?
Round your answer to the nearest cent.
$

Answer 1

To find the initial deposit that Marshall made in his savings account, we can use the formula for continuous compounding, which is given by:

\[
A = Pe^{rt}
\]

Where:
- \( A \) is the amount of money accumulated after time \( t \) (the current balance),
- \( P \) is the principal amount (the initial deposit),
- \( r \) is the annual interest rate (as a decimal),
- \( t \) is the time in years,
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).

Given:
- \( A = 200 \) (the current balance),
- \( r = 0.01 \) (1% interest rate),
- \( t = 9 \) (the time in years).

We need to solve for \( P \):

\[
200 = P e^{0.01 \times 9}
\]

Calculating \( e^{0.01 \times 9} \):

\[
0.01 \times 9 = 0.09
\]
\[
e^{0.09} \approx 1.09417
\]

Now, substituting \( e^{0.09} \) back into the formula:

\[
200 = P \cdot 1.09417
\]

Now, solve for \( P \):

\[
P = \frac{200}{1.09417} \approx 182.06
\]

Therefore, the initial deposit that Marshall made is approximately:

\[
\boxed{182.06}
\]