Solve the problem.
-Find out how long it takes a $\$ 3000$ investment to double if it is invested at $7 \%$ compounded monthly. Round to the nearest tenth of a year. Use the formula $\mathrm { A } = \mathrm { P } \left( 1 + \frac { \mathrm { r } } { \mathrm { n } } \right) ^ { \mathrm { nt } }$.
A) $9.9$ years
B) $10.1$ years
C) $9.7$ years
D) $10.3$ years

Question

Quizplus · Accepted Answer

To find the time it takes for an investment to double at a 7% annual interest rate compounded monthly, we use the formula $A = P\left(1 + \frac{r}{n}\right)^{nt}$, 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. Since we want the investment to double, $A = 2P = 2 \times 3000 = 6000$, $r = 0.07$ (7% as a decimal), and $n = 12$ (since the interest is compounded monthly).Substituting the values into the formula, we get:$$6000 = 3000\left(1 + \frac{0.07}{12}\right)^{12t}$$Dividing both sides by 3000, we get:$$2 = \left(1 + \frac{0.07}{12}\right)^{12t}$$Taking the natural logarithm of both sides:$$\ln(2) = \ln\left(\left(1 + \frac{0.07}{12}\right)^{12t}\right)$$Using the property of logarithms, we can bring down the exponent:$$12t\ln\left(1 + \frac{0.07}{12}\right) = \ln(2)$$Solving for $t$:$$t = \frac{\ln(2)}{12\ln\left(1 + \frac{0.07}{12}\right)}$$Using a calculator:$$t \approx 9.9$$Therefore, it takes approximately 9.9 years for the investment to double, making choice A correct.

Solve the Problem $3000\$ 3000$3000 Investment to Double If It Is Invested At

Solve the Problem $\$ 3000$ Investment to Double If It Is Invested At