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?\n\ncompound interest formula: $v(t)=p(1 + \\frac{r}{n})^{nt}$\n\nt = years since initial deposit\nn = number of times compounded per year\nr = annual interest rate (as a decimal)\np = initial (principal) investment\nv(t) = value of investment after t years\n\n2 years\n9 years\n18 years\n36 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?\n\ncompound interest formula: $v(t)=p(1 + \\frac{r}{n})^{nt}$\n\nt = years since initial deposit\nn = number of times compounded per year\nr = annual interest rate (as a decimal)\np = initial (principal) investment\nv(t) = value of investment after t years\n\n2 years\n9 years\n18 years\n36 years

Answer

Answer:

C. 18 years

Explanation:

Step1: Identify the values

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

Step2: Substitute into formula

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

Step3: Simplify the equation

$\frac{3875.79}{1900}=(1 + 0.02)^{2t}$ $2.04 = 1.02^{2t}$

Step4: Take the natural - log of both sides

$\ln(2.04)=\ln(1.02^{2t})$ Using the property $\ln(a^{b})=b\ln(a)$, we get $\ln(2.04)=2t\ln(1.02)$

Step5: Solve for t

$t=\frac{\ln(2.04)}{2\ln(1.02)}$ $\ln(2.04)\approx0.712$, $\ln(1.02)\approx0.0198$ $t=\frac{0.712}{2\times0.0198}=\frac{0.712}{0.0396}\approx18$ years