$\$ 6500$ is invested at $10 \%$ compounded quarterly. In how many years will the account have grown to $\$ 11,500$ ? Round your answer to the nearest tenth of a year.
A) 1.1 years
B) 5.8 years
C) 10.6 years
D) 6.0 years

Question

Quizplus · Accepted Answer

The formula for compound interest is $A = P(1 + \frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after n years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (decimal), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years. Here, $A = \$11,500$, $P = \$6,500$, $r = 10\% = 0.10$, and $n = 4$ (since the interest is compounded quarterly). We solve for $t$:$$\$11,500 = \$6,500 \left(1 + \frac{0.10}{4}\right)^{4t}$$$$\frac{\$11,500}{\$6,500} = \left(1 + \frac{0.10}{4}\right)^{4t}$$$$1.76923 \approx (1.025)^{4t}$$Taking the natural logarithm of both sides:$$\ln(1.76923) \approx 4t \ln(1.025)$$$$t \approx \frac{\ln(1.76923)}{4\ln(1.025)}$$$$t \approx 5.8 \text{ years}$$Therefore, the account will have grown to \$11,500 in approximately 5.8 years.

$6500\$ 6500$6500 Is Invested At 10%10 \%10% Compounded Quarterly $11,500\$ 11,500$11,500 ? Round Your Answer to the Nearest Tenth of a How

$\$ 6500$ Is Invested At $10 \%$ Compounded Quarterly $\$ 11,500$ ? Round Your Answer to the Nearest Tenth of a How