Question

a person invests 5500 dollars in the bank. the bank pays 4.25% interest compounded quarterly. to the nearest tenth of a year, how long must a person leave the money in the bank until it teaches 8600 dollars?

Answer 1

Step 1. The information that we have is:

-The initial investment, which will be the principal P:

$P=5,500$

-The interest rate r which we will represent as a decimal number:

$\begin{gathered} r=4.25/100 \\ r=0.0425 \end{gathered}$

-The investment is compounded quarterly, this is 4 times per year:

$n=4$

Required: Find the time in years it will take for the amount to be $8,600.

This final amount is A:

$A=8,600$

Step 2. Once we have defined all of our values, we use the compound interest formula:

$A=P(1+\frac{r}{n})^{nt}$

And substitute the known values:

$8,600=5,500(1+\frac{0.0425}{4})^{4t}$

Step 3. To simplify, we solve the operations in the pair of parentheses:

$\begin{gathered} 8,600=5,500(1+0.010625)^{4t} \\ \downarrow \\ 8,600=5,500(1.010625)^{4t} \end{gathered}$

Then, divide both sides by 5,500:

$\begin{gathered} \frac{8,600}{5,500}=\frac{5,500(1.010625)^{4t}}{5,500} \\ \downarrow \\ 1.563636364=(1.010625)^{4t} \end{gathered}$

Step 4. Since we need to find the value of t which is in the exponent of the equation, we apply the logarithm to both sides of the equation:

$log(1.563636364)=log(1.010625)^{4t}$

And due to the following property of logarithms:

$log(x^n)=nlog(x)$

The expression can be written as follows:

$log(1.563636364)=4t\times log(1.010625)$

Step 5. Solving for t:

$\begin{gathered} 4t=\frac{log(1.563636364)}{log(1.010625)} \\ \downarrow \\ 4t=42.29502968 \\ \downarrow \\ t=\frac{42.29502968}{4} \\ \downarrow \\ t=10.57375742 \end{gathered}$

Rounding the time to the nearest tenth:

$t=10.6$

The time is 10.6 years.

Answer:

10.6 years