Example

Try It 10.87: Evaluating an Exponential Growth Model

Practice solving an exponential growth problem by finding the growth rate and using it to predict a future amount. Suppose a bacteria population grows from 100100 to 500500 in 66 hours. Using the exponential growth formula A=A0ektA = A_0 e^{kt}, substitute the values to find the rate: 500=100ek6500 = 100 e^{k \cdot 6}. Dividing by 100100 gives 5=e6k5 = e^{6k}. Applying the natural logarithm yields ln5=6k\ln 5 = 6k, so the growth rate is k=ln56k = \frac{\ln 5}{6}. To find the population after 2424 hours, substitute the rate and time into the formula: A=100eln5624A = 100 e^{\frac{\ln 5}{6} \cdot 24}. Simplifying the exponent (ln5624=4ln5\frac{\ln 5}{6} \cdot 24 = 4 \ln 5) gives A=100e4ln5A = 100 e^{4 \ln 5}, which evaluates to 62,50062{,}500. Therefore, there will be 62,50062{,}500 bacteria.

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Updated 2026-05-26

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