To find the period [tex]\( t \)[/tex] in years that David's investment of \[tex]$230 will reach at least \$[/tex]415 with an interest rate of 3%, we use the compound interest formula:
[tex]\[ A = P(1 + r)^t \][/tex]
Here,
- [tex]\( A \)[/tex] is the amount after [tex]\( t \)[/tex] years (\[tex]$415),
- \( P \) is the initial investment (\$[/tex]230),
- [tex]\( r \)[/tex] is the annual interest rate (0.03),
- [tex]\( t \)[/tex] is the time period in years.
First, we rewrite the formula to solve for [tex]\( t \)[/tex]:
[tex]\[ \frac{A}{P} = (1 + r)^t \][/tex]
[tex]\[ \frac{415}{230} = (1 + 0.03)^t \][/tex]
[tex]\[ 1.804347826087 = 1.03^t \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm (ln) on both sides:
[tex]\[ \ln(1.804347826087) = \ln(1.03^t) \][/tex]
Using the logarithm power rule [tex]\( \ln(a^b) = b\ln(a) \)[/tex], we get:
[tex]\[ \ln(1.804347826087) = t \cdot \ln(1.03) \][/tex]
[tex]\[ t = \frac{\ln(1.804347826087)}{\ln(1.03)} \][/tex]
Using a calculator to find the logarithms:
[tex]\[ t = \frac{\ln(1.804347826087)}{\ln(1.03)} \approx 19.96695287192616 \][/tex]
Rounding this to the nearest year:
[tex]\[ t \approx 20 \][/tex]
Thus, the value of David's investment will be at least \$415 after a period of [tex]\( 20 \)[/tex] years.
