To determine how long it will take for a [tex]$2000 investment to grow to $[/tex]3212 with an annual interest rate of 6%, compounded monthly, you can use the formula for compound interest:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A \)[/tex] is the final amount ([tex]$3212),
- \( P \) is the principal or initial amount ($[/tex]2000),
- [tex]\( r \)[/tex] is the annual interest rate (0.06),
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year (12), and
- [tex]\( t \)[/tex] is the time in years.
First, we solve for [tex]\( t \)[/tex]. Rearrange the formula to solve for [tex]\( t \)[/tex]:
[tex]\[ 3212 = 2000 \left(1 + \frac{0.06}{12}\right)^{12t} \][/tex]
1. Divide both sides by 2000:
[tex]\[ \frac{3212}{2000} = \left(1 + \frac{0.06}{12}\right)^{12t} \][/tex]
2. Simplify the fraction:
[tex]\[ 1.606 = \left(1 + \frac{0.06}{12}\right)^{12t} \][/tex]
3. Calculate the monthly interest rate:
[tex]\[ 1 + \frac{0.06}{12} = 1 + 0.005 = 1.005 \][/tex]
4. Substitute 1.005 into the equation:
[tex]\[ 1.606 = 1.005^{12t} \][/tex]
5. Take the natural logarithm (ln) of both sides to solve for the exponent:
[tex]\[ \ln(1.606) = \ln(1.005^{12t}) \][/tex]
6. Use the logarithm power rule [tex]\(\ln(a^b) = b \ln(a)\)[/tex]:
[tex]\[ \ln(1.606) = 12t \ln(1.005) \][/tex]
7. Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(1.606)}{12 \ln(1.005)} \][/tex]
8. Compute the values:
[tex]\[ \ln(1.606) \approx 0.4730 \][/tex]
[tex]\[ \ln(1.005) \approx 0.0049875 \][/tex]
9. Substitute the values into the equation:
[tex]\[ t = \frac{0.4730}{12 \times 0.0049875} \][/tex]
[tex]\[ t = \frac{0.4730}{0.05985} \][/tex]
[tex]\[ t \approx 7.9027 \][/tex]
Rounding to the nearest hundredth:
[tex]\[ t \approx 7.92 \][/tex]
So, it will take approximately 7.92 years for the [tex]$2000 investment to grow to $[/tex]3212 at an annual rate of 6%, compounded monthly.
