To determine how many years it will take for \($6,000 to grow to $\)8,000 with continuous compounding at an annual interest rate of 9.0%, we can use the formula for continuous compounding:
\(\[ A = P \cdot e^{rt} \]\)
where:
- \(\( A \)\) is the final amount (\($8,000),
- \( P \) is the principal amount ($\)6,000),
- \(\( r \)\) is the annual interest rate (0.09),
- \(\( t \)\) is the time in years.
First, let's rewrite the formula to solve for \(\( t \)\):
\(\[ \frac{A}{P} = e^{rt} \]\)
Taking the natural logarithm of both sides, we get:
\(\[ \ln\left(\frac{A}{P}\right) = rt \]\)
So,
\(\[ t = \frac{\ln\left(\frac{A}{P}\right)}{r} \]\)
Now substitute the given values into the formula:
\(\[ t = \frac{\ln\left(\frac{8000}{6000}\right)}{0.09} \]\)
Simplify the fraction inside the natural logarithm:
\(\[ \frac{8000}{6000} = \frac{4}{3} \]\)
So we have:
\(\[ t = \frac{\ln\left(\frac{4}{3}\right)}{0.09} \]\)
Using a calculator to find the natural logarithm:
\(\[ \ln\left(\frac{4}{3}\right) \approx 0.2877 \]\)
Now plug this back into the formula:
\(\[ t = \frac{0.2877}{0.09} \approx 3.1967 \]\)
Thus, it will take approximately 3.2 years for \($6,000 to grow to $\)8,000 at an annual interest rate of 9.0% compounded continuously.
The correct answer, closest to 3.2 years, is:
OB. 3 years
