Select the correct answer.
How many years (approx) will it take [tex]$6,000 to amount to $[/tex]8,000 if it is invested at an annual rate of 9.0% compounded continuously?
OA.
23 years
OB.
3 years
O c. 9 years
OD. 2 years
O E.
42 years



Answer :

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