a sum of money is invested at 12% compounded quarterly. about how long will it take for the amount of money…

a sum of money is invested at 12% compounded quarterly. about how long will it take for the amount of money to double?\ncompound interest formula: $v(t)=p(1 + \\frac{r}{n})^{nt}$\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\n5.9 years\n6.1 years\n23.4 years\n24.5 years

a sum of money is invested at 12% compounded quarterly. about how long will it take for the amount of money to double?\ncompound interest formula: $v(t)=p(1 + \\frac{r}{n})^{nt}$\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\n5.9 years\n6.1 years\n23.4 years\n24.5 years

Answer

Explanation:

Step1: Identify given values

$r = 0.12$, $n=4$, let $P$ be the initial investment and $V(t)=2P$.

Step2: Substitute into compound - interest formula

$2P=P\left(1 +\frac{0.12}{4}\right)^{4t}$. Divide both sides by $P$ to get $2=(1 + 0.03)^{4t}$, so $2 = 1.03^{4t}$.

Step3: Take the natural logarithm of both sides

$\ln(2)=\ln(1.03^{4t})$. Using the property $\ln(a^{b})=b\ln(a)$, we have $\ln(2)=4t\ln(1.03)$.

Step4: Solve for $t$

$t=\frac{\ln(2)}{4\ln(1.03)}$. Calculate $\ln(2)\approx0.6931$ and $\ln(1.03)\approx0.0296$. Then $t=\frac{0.6931}{4\times0.0296}=\frac{0.6931}{0.1184}\approx5.9$.

Answer:

5.9 years