Solve the problem.
-Eight million E. coli bacteria are present in a laboratory culture. An antibacterial agent is introduced and the population of bacteria $P ( t )$ decreases by half every $5 \mathrm { hr }$. The population can be represented by $P ( t ) = 8,000,000 \left( \frac { 1 } { 2 } \right) ^ { t / 5 }$. Convert this to an exponential function using base $e$.
A) $P ( t ) = 8,000,000 e ^ { - 0.13863 t }$
B) $P ( t ) = 4,000,000 e ^ { - 0.13863 t }$
C) $P ( t ) = 8,000,000 e ^ { - 0.1 t }$
D) $P ( t ) = 4,000,000 e ^ { - 0.1 t }$

Question

Quizplus · Accepted Answer

The given function $P(t) = 8,000,000 \left( \frac{1}{2} \right)^{t/5}$ can be converted to an exponential function with base $e$ by using the identity $a^x = e^{x \ln(a)}$. Here, $a = \frac{1}{2}$ and $x = \frac{t}{5}$, so the function becomes $P(t) = 8,000,000 e^{\frac{t}{5} \ln(\frac{1}{2})}$. Since $\ln(\frac{1}{2})$ is approximately $-0.693147$, the exponent becomes $\frac{t}{5} \times -0.693147 = -0.13863t$, resulting in $P(t) = 8,000,000 e^{-0.13863t}$.

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