Answer :
To find out after how many years a loan of [tex]$38,000 compounded annually at an interest rate of 5.5% will reach or exceed $[/tex]83,000, we use the compound interest formula:
[tex]\[ A = P(1 + r)^t \][/tex]
Where:
- [tex]\( A \)[/tex] is the amount due, which is [tex]$83,000. - \( P \) is the principal amount, which is $[/tex]38,000.
- [tex]\( r \)[/tex] is the annual interest rate, expressed as a decimal, so 5.5% becomes 0.055.
- [tex]\( t \)[/tex] is the number of years.
We need to solve for [tex]\( t \)[/tex]. First, let's plug the given values into the formula:
[tex]\[ 83000 = 38000 \times (1 + 0.055)^t \][/tex]
Next, we will isolate [tex]\( (1 + 0.055)^t \)[/tex]:
[tex]\[ \frac{83000}{38000} = (1.055)^t \][/tex]
Simplify the left side of the equation:
[tex]\[ \frac{83000}{38000} \approx 2.1842 \][/tex]
So, the equation now is:
[tex]\[ 2.1842 = (1.055)^t \][/tex]
To solve for [tex]\( t \)[/tex], we use logarithms. Taking the natural logarithm (ln) of both sides gives:
[tex]\[ \ln(2.1842) = \ln((1.055)^t) \][/tex]
Using the logarithm power rule:
[tex]\[ \ln(2.1842) = t \cdot \ln(1.055) \][/tex]
Now we solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(2.1842)}{\ln(1.055)} \][/tex]
Using a calculator to find the values of the logarithms:
[tex]\[ \ln(2.1842) \approx 0.7819 \][/tex]
[tex]\[ \ln(1.055) \approx 0.0539 \][/tex]
So,
[tex]\[ t = \frac{0.7819}{0.0539} \approx 14.5 \][/tex]
Since we need the smallest possible whole number answer and we are looking for the time when the amount reaches or exceeds [tex]$83,000, we round up to the next whole number because even if \( t \) is slightly over 14.5 years, it will need the full next year to complete the amount. Therefore, the number of years it will take for the loan to reach or exceed $[/tex]83,000 is [tex]\(\boxed{15}\)[/tex].
[tex]\[ A = P(1 + r)^t \][/tex]
Where:
- [tex]\( A \)[/tex] is the amount due, which is [tex]$83,000. - \( P \) is the principal amount, which is $[/tex]38,000.
- [tex]\( r \)[/tex] is the annual interest rate, expressed as a decimal, so 5.5% becomes 0.055.
- [tex]\( t \)[/tex] is the number of years.
We need to solve for [tex]\( t \)[/tex]. First, let's plug the given values into the formula:
[tex]\[ 83000 = 38000 \times (1 + 0.055)^t \][/tex]
Next, we will isolate [tex]\( (1 + 0.055)^t \)[/tex]:
[tex]\[ \frac{83000}{38000} = (1.055)^t \][/tex]
Simplify the left side of the equation:
[tex]\[ \frac{83000}{38000} \approx 2.1842 \][/tex]
So, the equation now is:
[tex]\[ 2.1842 = (1.055)^t \][/tex]
To solve for [tex]\( t \)[/tex], we use logarithms. Taking the natural logarithm (ln) of both sides gives:
[tex]\[ \ln(2.1842) = \ln((1.055)^t) \][/tex]
Using the logarithm power rule:
[tex]\[ \ln(2.1842) = t \cdot \ln(1.055) \][/tex]
Now we solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(2.1842)}{\ln(1.055)} \][/tex]
Using a calculator to find the values of the logarithms:
[tex]\[ \ln(2.1842) \approx 0.7819 \][/tex]
[tex]\[ \ln(1.055) \approx 0.0539 \][/tex]
So,
[tex]\[ t = \frac{0.7819}{0.0539} \approx 14.5 \][/tex]
Since we need the smallest possible whole number answer and we are looking for the time when the amount reaches or exceeds [tex]$83,000, we round up to the next whole number because even if \( t \) is slightly over 14.5 years, it will need the full next year to complete the amount. Therefore, the number of years it will take for the loan to reach or exceed $[/tex]83,000 is [tex]\(\boxed{15}\)[/tex].
