An amount of $4,400 is invested in an account with annually compounded interest. After 5 years the account holds $5,790. What was the annual interest rate?

Step-by-step solution

1. 1

   Start from the compound interest formula. For interest compounded once per year,

   $A=P(1+r)^t$

   where $P=4400$ is the principal, $A=5790$ is the balance after $t=5$ years, and $r$ is the annual rate as a decimal. Everything is known except $r$, so the problem is an algebra problem: isolate $r$.

2. 2

   Isolate the growth factor. Substituting the values and dividing both sides by the principal removes $P$:

   $5790=4400(1+r)^5\ \Longrightarrow\ (1+r)^5=\frac{5790}{4400}=1.3159090909\ldots$

   This number is the total growth over five years — the money grew by about $31.6\%$ in total, not per year.

3. 3

   Undo the exponent with a fifth root. Raising both sides to the power $\tfrac15$ converts total growth into annual growth:

   $1+r=\left(1.3159090909\right)^{1/5}=1.05644083.$

   A common mistake here is to divide the total growth by $5$ instead; that would give about $6.3\%$ and is wrong, because compounding is multiplicative, not additive.

4. 4

   Subtract 1 and convert to a percentage.

   $r=1.05644083-1=0.05644083=5.644083\%.$

5. 5

   Round and verify. To one decimal place the rate is $\mathbf{5.6\%}$. Checking both candidate roundings against the target balance: 4400(1.056)^5=\5{,}777.93$and$4400(1.057)^5=$5{,}805.34$, while the unrounded rate reproduces$$5{,}790.00$exactly. Since$5.644%$is below the$5.65%$midpoint, it rounds **down** to$5.6%$, and$5.6%$ is indeed the closer of the two.