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? 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 5.9 years 6.1 years 23.4 years 24.5 years

a sum of money is invested at 12% compounded quarterly. about how long will it take for the amount of money to double? 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 5.9 years 6.1 years 23.4 years 24.5 years

Answer

Explanation:

Step1: Identify values

Given $r = 0.12$ (12% as a decimal), $n=4$ (compounded quarterly), and $V(t) = 2P$. Substitute into compound - interest formula $V(t)=P(1 +\frac{r}{n})^{nt}$. $2P=P(1+\frac{0.12}{4})^{4t}$

Step2: Simplify the equation

Divide both sides of the equation by $P$ (since $P\neq0$). We get $2=(1 + 0.03)^{4t}$, which simplifies to $2=(1.03)^{4t}$.

Step3: Take the natural logarithm of both sides

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

Step4: Solve for $t$

First, find $\ln(2)\approx0.6931$ and $\ln(1.03)\approx0.0296$. Then $t=\frac{\ln(2)}{4\ln(1.03)}$. $t=\frac{0.6931}{4\times0.0296}=\frac{0.6931}{0.1184}\approx5.86\approx5.9$

Answer:

5.9 years