To determine the length of time the money was invested, we need to use the compound interest formula. Given that the interest is compounded annually, the formula we will use is:
[tex]\[
A = P(1 + r)^t
\][/tex]
where:
- \( A \) is the final amount of money (the balance after interest is applied),
- \( P \) is the principal amount (the initial amount of money),
- \( r \) is the annual interest rate (as a decimal),
- \( t \) is the number of years the money is invested.
We are given:
- \( P = 288 \) (the initial principal),
- \( r = 0.07 \) (the annual interest rate as a decimal),
- \( A = 581.23 \) (the final balance).
We need to solve for \( t \).
First, we start by rearranging the compound interest formula to solve for \( t \):
[tex]\[
\frac{A}{P} = (1 + r)^t
\][/tex]
Next, we take the natural logarithm (ln) of both sides to get:
[tex]\[
\ln\left(\frac{A}{P}\right) = t \cdot \ln(1 + r)
\][/tex]
Now, solve for \( t \):
[tex]\[
t = \frac{\ln\left(\frac{A}{P}\right)}{\ln(1 + r)}
\][/tex]
Let's substitute the given values into the formula:
[tex]\[
t = \frac{\ln\left(\frac{581.23}{288}\right)}{\ln(1 + 0.07)}
\][/tex]
Calculating the values inside the logarithm:
[tex]\[
\frac{581.23}{288} \approx 2.018826
\][/tex]
Now, plug this into the formula:
[tex]\[
t = \frac{\ln(2.018826)}{\ln(1.07)}
\][/tex]
To find the natural logarithm values:
[tex]\[
\ln(2.018826) \approx 0.700353
\][/tex]
[tex]\[
\ln(1.07) \approx 0.067658
\][/tex]
Finally, divide these values to find \( t \):
[tex]\[
t \approx \frac{0.700353}{0.067658} \approx 10.378
\][/tex]
So, the money was invested for approximately 10.38 years.
