To solve for the number of years [tex]\( t \)[/tex] that Jacques' money was in the account earning compound interest, we will use the compound interest formula:
[tex]\[ V(t) = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Given values from the problem are:
- [tex]\( P = 1900 \)[/tex] (initial principal investment)
- [tex]\( r = 0.04 \)[/tex] (annual interest rate as a decimal)
- [tex]\( n = 2 \)[/tex] (number of times interest is compounded per year, since it is compounded semiannually)
- [tex]\( V = 3875.79 \)[/tex] (value of investment after [tex]\( t \)[/tex] years)
We need to find [tex]\( t \)[/tex]. Let’s break down the problem step by step:
1. Set up the equation using the given values:
[tex]\[ 3875.79 = 1900 \left(1 + \frac{0.04}{2}\right)^{2t} \][/tex]
2. Simplify inside the parentheses:
[tex]\[ 1 + \frac{0.04}{2} = 1 + 0.02 = 1.02 \][/tex]
So the formula becomes:
[tex]\[ 3875.79 = 1900 \times (1.02)^{2t} \][/tex]
3. Divide both sides of the equation by 1900 to isolate the exponential term:
[tex]\[ \frac{3875.79}{1900} = (1.02)^{2t} \][/tex]
This gives us:
[tex]\[ 2.0398894736842106 = (1.02)^{2t} \][/tex]
4. Take the natural logarithm on both sides to solve for the exponent:
[tex]\[ \ln(2.0398894736842106) = \ln((1.02)^{2t}) \][/tex]
5. Use the property of logarithms to bring the exponent down:
[tex]\[ \ln(2.0398894736842106) = 2t \cdot \ln(1.02) \][/tex]
6. Solve for [tex]\( 2t \)[/tex] by dividing both sides by [tex]\(\ln(1.02)\)[/tex]:
[tex]\[ 2t = \frac{\ln(2.0398894736842106)}{\ln(1.02)} \][/tex]
[tex]\[ 2t = \frac{36.00005272832286}{\ln(1.02)} \][/tex]
[tex]\[ 2t = 36.00005272832286 \][/tex]
7. Finally, solve for [tex]\( t \)[/tex] by dividing by 2:
[tex]\[ t = \frac{36.00005272832286}{2} = 18.00002636416143 \][/tex]
So, the length of time that Jacques' money was in the account is approximately [tex]\( 18 \)[/tex] years.
Thus, the correct answer is [tex]\( \boxed{18 \text{ years}} \)[/tex].
