To determine how long it will take for a [tex]$3000 investment to grow to $[/tex]13,848 at an annual rate of 8%, with interest compounded semiannually, we will follow the compound interest formula and solve for the time.
The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A \)[/tex] is the final amount ([tex]$13,848).
- \( P \) is the principal amount ($[/tex]3000).
- [tex]\( r \)[/tex] is the annual interest rate (0.08).
- [tex]\( n \)[/tex] is the number of times interest is compounded per year (2, since it is semiannually).
- [tex]\( t \)[/tex] is the time in years that we need to find.
First, let's rearrange the formula to solve for [tex]\( t \)[/tex]:
[tex]\[ \left(1 + \frac{r}{n}\right)^{nt} = \frac{A}{P} \][/tex]
Take the natural logarithm (ln) of both sides:
[tex]\[ \ln\left(\left(1 + \frac{r}{n}\right)^{nt}\right) = \ln\left(\frac{A}{P}\right) \][/tex]
Use the properties of logarithms to bring the exponent [tex]\( nt \)[/tex] down:
[tex]\[ nt \cdot \ln\left(1 + \frac{r}{n}\right) = \ln\left(\frac{A}{P}\right) \][/tex]
Now, solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln\left(\frac{A}{P}\right)}{n \cdot \ln\left(1 + \frac{r}{n}\right)} \][/tex]
Substitute the values [tex]\( A = 13848 \)[/tex], [tex]\( P = 3000 \)[/tex], [tex]\( r = 0.08 \)[/tex], and [tex]\( n = 2 \)[/tex]:
[tex]\[ t = \frac{\ln\left(\frac{13848}{3000}\right)}{2 \cdot \ln\left(1 + \frac{0.08}{2}\right)} \][/tex]
Calculate the numerator:
[tex]\[ \ln\left(\frac{13848}{3000}\right) = \ln(4.616) \][/tex]
Calculate the denominator:
[tex]\[ 2 \cdot \ln\left(1 + \frac{0.08}{2}\right) = 2 \cdot \ln(1.04) \][/tex]
Putting this into the fraction:
[tex]\[ t = \frac{\ln(4.616)}{2 \cdot \ln(1.04)} \][/tex]
Using the natural logarithm values:
[tex]\[ \ln(4.616) \approx 1.529 \][/tex]
[tex]\[ \ln(1.04) \approx 0.0392 \][/tex]
Calculate the denominator:
[tex]\[ 2 \cdot 0.0392 = 0.0784 \][/tex]
Now, divide the values:
[tex]\[ t = \frac{1.529}{0.0784} \approx 19.50 \][/tex]
Therefore, the number of years it will take for the investment to grow to $13,848 is approximately 19.50 years, rounded to the nearest hundredth.
