jacques deposited $1,900 into an account that earns 4% interest compounded semiannually. after t years…

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: $v(t)=p(1 + \frac{r}{n})^{nt}$ 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 18 years 2 years 9 years 36 years

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: $v(t)=p(1 + \frac{r}{n})^{nt}$ 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 18 years 2 years 9 years 36 years

Answer

Explanation:

Step1: Identify the values

$P = 1900$, $r=0.04$, $n = 2$, $V(t)=3875.79$

Step2: Substitute into formula

$3875.79=1900(1 +\frac{0.04}{2})^{2t}$

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 $\ln(a^{b})=b\ln(a)$, 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