1Cademy - Try It 10.85 and 10.86: Solving for the Rate of Continuous Compound Interest

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Example 10.43: Solving for the Rate of Continuous Compound Interest

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Try It 10.85 and 10.86: Solving for the Rate of Continuous Compound Interest

Practice finding the required interest rate for a continuously compounded investment using the formula $A = Pe^{rt}$ . For an initial investment of 10{,}000 at age $21$ that needs to grow to 150{,}000 by age $50$ (a period of $29$ years), substitute the values to get $150{,}000 = 10{,}000e^{29r}$ . Dividing by 10{,}000 yields $15 = e^{29r}$ . Taking the natural logarithm of both sides gives $\ln 15 = 29r$ , resulting in $r = \frac{\ln 15}{29} \approx 0.093$ , or approximately 9.3%. Similarly, for a 15{,}000 investment at age $25$ that must reach 90{,}000 by age $40$ ( $15$ years), the equation is $90{,}000 = 15{,}000e^{15r}$ . Dividing by 15{,}000 gives $6 = e^{15r}$ . Applying the natural logarithm yields $\ln 6 = 15r$ , so $r = \frac{\ln 6}{15} \approx 0.119$ , which is about 11.9%.

Updated 2026-06-03

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OpenStax Intermediate Algebra

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