Compound Interest Math Example 5

Follow the full solution, then compare it with the other examples linked below.

Example 5

hard
What annual interest rate, compounded monthly, is needed to grow \3{,}000to to \4,5004{,}500 in 6 years?

Solution

  1. 1
    From 4500=3000(1+r/12)724500 = 3000(1 + r/12)^{72}, divide: (1+r/12)72=1.5(1 + r/12)^{72} = 1.5.
  2. 2
    Take the 72nd root: 1+r/12=1.51/72โ‰ˆ1.0056541 + r/12 = 1.5^{1/72} \approx 1.005654.
  3. 3
    Solve: r/12โ‰ˆ0.005654r/12 \approx 0.005654, so rโ‰ˆ0.06785โ‰ˆ6.79%r \approx 0.06785 \approx 6.79\%.

Answer

rโ‰ˆ6.79%r \approx 6.79\%
To find the rate, isolate the base of the exponential expression, then use roots or logarithms. This is the reverse of the typical compound interest calculation.

About Compound Interest

Interest calculated on both the initial principal and the accumulated interest from previous periods. The formula A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt} gives the amount after tt years, and A=PertA = Pe^{rt} gives the continuously compounded amount.

Learn more about Compound Interest โ†’

More Compound Interest Examples

Example 1 easy

You invest [formula]5{,}000[formula]6%$ annual interest compounded quarterly. Find the amount after

Example 2 medium

How long does it take for an investment to double at [formula] annual interest compounded monthly?

Example 3 medium

Find the amount when [formula]2{,}000[formula]5%$ compounded continuously for 4 years.

Example 4 easy

You deposit [formula]1{,}200[formula]4%$ compounded annually. What is the balance after 5 years?

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