Answer :
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.
[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.
