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