jacques deposited $1,900 into an account that...
jacques deposited $1,900 into an account that earns 4% interest compounded semiannually. after t years, jacques has $3,875.79 in the account. assuming he made no additional deposits or withdrawals, how long was the money in the account? compound - interest formula: v(t)=p(1 + r/n)^(nt) t = years since initial deposit n = number of times compounded per year r = annual interest rate (as a decimal) p = initial (principal) investment v(t) = value of investment after t years 2 years 9 years 18 years 36 years
Answer
# Explanation:
## Step1: Identify the values
$P = 1900$, $V(t)=3875.79$, $r = 0.04$, $n = 2$
## Step2: Substitute into compound - interest formula
$3875.79=1900\left(1 +\frac{0.04}{2}\right)^{2t}$
## Step3: Simplify the equation
First, simplify $1+\frac{0.04}{2}=1 + 0.02=1.02$. So the equation becomes $\frac{3875.79}{1900}=1.02^{2t}$. And $\frac{3875.79}{1900}=2.03989$. So $2.03989 = 1.02^{2t}$.
## Step4: Take the natural logarithm of both sides
$\ln(2.03989)=\ln(1.02^{2t})$. Using the property of logarithms $\ln(a^{b})=b\ln(a)$, we get $\ln(2.03989)=2t\ln(1.02)$.
## Step5: Solve for $t$
$t=\frac{\ln(2.03989)}{2\ln(1.02)}$. Calculate $\ln(2.03989)\approx0.712$ and $\ln(1.02)\approx0.0198$. Then $2\ln(1.02)\approx0.0396$. So $t=\frac{0.712}{0.0396}\approx18$.
# Answer:
18 years