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 2026-05-25

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OpenStax Intermediate Algebra

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