To determine the number of years [tex]\( t \)[/tex] it takes for an investment to double when continuously compounded at an annual interest rate of [tex]\( 8\% \)[/tex], we can use the formula for continuous compounding:
[tex]\[ A = Pe^{rt} \][/tex]
where:
- [tex]\( P \)[/tex] is the principal amount (initial investment),
- [tex]\( r \)[/tex] is the annual interest rate (as a decimal),
- [tex]\( t \)[/tex] is the time in years,
- [tex]\( A \)[/tex] is the accumulated amount after time [tex]\( t \)[/tex],
- [tex]\( e \)[/tex] is the base of the natural logarithm.
Given:
- Principal amount [tex]\( P = 9500 \)[/tex] dollars,
- Annual interest rate [tex]\( r = 8\% = 0.08 \)[/tex],
- The accumulated amount [tex]\( A \)[/tex] is double the principal, so [tex]\( A = 2P = 2 \times 9500 = 19000 \)[/tex] dollars.
The continuous compounding formula is:
[tex]\[ A = P e^{rt} \][/tex]
Plugging in the known values:
[tex]\[ 19000 = 9500 e^{0.08t} \][/tex]
Next, we need to solve for [tex]\( t \)[/tex].
1. Divide both sides by 9500:
[tex]\[ \frac{19000}{9500} = e^{0.08t} \][/tex]
This simplifies to:
[tex]\[ 2 = e^{0.08t} \][/tex]
2. Take the natural logarithm (ln) of both sides to eliminate the base [tex]\( e \)[/tex]:
[tex]\[ \ln(2) = \ln(e^{0.08t}) \][/tex]
By the properties of logarithms, the right-hand side simplifies to:
[tex]\[ \ln(2) = 0.08t \][/tex]
3. Solve for [tex]\( t \)[/tex] by dividing both sides by 0.08:
[tex]\[ t = \frac{\ln(2)}{0.08} \][/tex]
Using a calculator to find the natural logarithm of 2:
[tex]\[ \ln(2) \approx 0.693147 \][/tex]
So,
[tex]\[ t = \frac{0.693147}{0.08} \approx 8.6643 \][/tex]
Finally, rounding to one decimal place:
[tex]\[ t \approx 8.7 \][/tex]
Therefore, it takes approximately [tex]\( 8.7 \)[/tex] years for the investment to double when compounded continuously at an annual rate of [tex]\( 8\% \)[/tex].
