Question

jaxon invested $960 in an account paying an interest rate of 6.1% compounded monthly. assuming no deposits or withdrawals are made, how long would it take, to the nearest year, for the value of the account to reach $1,720?

Explanation:

Step1: Recall compound interest formula

The compound interest formula 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.
• $t$ is the time the money is invested for in years.

Given:

• $A = 1720$
• $P = 960$
• $r = 6.1\% = 0.061$
• $n = 12$ (compounded monthly)

We need to solve for $t$.

Step2: Substitute the given values into the formula

Substitute the values into the formula:
$$1720 = 960(1 + \frac{0.061}{12})^{12t}$$

Step3: Divide both sides by 960

$$\frac{1720}{960} = (1 + \frac{0.061}{12})^{12t}$$
Simplify $\frac{1720}{960} \approx 1.7917$ and $1 + \frac{0.061}{12} \approx 1 + 0.005083 = 1.005083$:
$$1.7917 = (1.005083)^{12t}$$

Step4: Take the natural logarithm of both sides

Take the natural logarithm (ln) of both sides:
$$\ln(1.7917) = \ln((1.005083)^{12t})$$
Using the logarithm power rule $\ln(a^b) = b\ln(a)$:
$$\ln(1.7917) = 12t \cdot \ln(1.005083)$$

Step5: Solve for t

First, calculate $\ln(1.7917) \approx 0.583$ and $\ln(1.005083) \approx 0.00507$:
$$0.583 = 12t \cdot 0.00507$$
First, solve for $12t$:
$$12t = \frac{0.583}{0.00507} \approx 115$$
Then, solve for $t$:
$$t = \frac{115}{12} \approx 9.58$$
To the nearest year, $t \approx 10$.

Answer:

10 years