To determine how long it will take for the value of a home purchased for [tex]$415,000 to reach $[/tex]530,000 given an annual increase rate of 5.2%, we can use the formula for compound interest, which is:
[tex]\[ A = P \times (1 + r)^n \][/tex]
where:
- [tex]\( A \)[/tex] is the final amount (in this case, [tex]$530,000),
- \( P \) is the initial principal balance (in this case, $[/tex]415,000),
- [tex]\( r \)[/tex] is the annual increase rate (in this case, 0.052),
- [tex]\( n \)[/tex] is the number of years.
We need to solve for [tex]\( n \)[/tex], so we can rearrange the formula as follows:
[tex]\[ (1 + r)^n = \frac{A}{P} \][/tex]
Taking the natural logarithm of both sides to isolate [tex]\( n \)[/tex]:
[tex]\[ \log((1 + r)^n) = \log\left(\frac{A}{P}\right) \][/tex]
Using the property of logarithms that [tex]\(\log(a^b) = b \log(a)\)[/tex], this simplifies to:
[tex]\[ n \log(1 + r) = \log\left(\frac{A}{P}\right) \][/tex]
Finally, solving for [tex]\( n \)[/tex]:
[tex]\[ n = \frac{\log\left(\frac{A}{P}\right)}{\log(1 + r)} \][/tex]
Let's substitute in the given values:
- [tex]\( A = 530,000 \)[/tex]
- [tex]\( P = 415,000 \)[/tex]
- [tex]\( r = 0.052 \)[/tex]
So,
[tex]\[ n = \frac{\log\left(\frac{530,000}{415,000}\right)}{\log(1 + 0.052)} \][/tex]
To find the values inside the logarithms:
[tex]\[ \frac{530,000}{415,000} \approx 1.277 \][/tex]
Thus, we have:
[tex]\[ n = \frac{\log(1.277)}{\log(1.052)} \][/tex]
Calculating the logarithms:
[tex]\[
\log(1.277) \approx 0.1068
\][/tex]
[tex]\[
\log(1.052) \approx 0.0216
\][/tex]
Thus:
[tex]\[ n = \frac{0.1068}{0.0216} \approx 4.825 \][/tex]
So, it will take approximately 4.825 years for the value of the home to reach $530,000.
The final answer is:
[tex]\[
n \approx 4.825
\][/tex]
