QuestionJuly 30, 2025

In a petri dish there are initially 2520 bacteria After 2 hours there are 5040 bacteria.How many bacteria will there be after 12 hours? Let's begin by defining our variables.What is P? P=[?],t=[ ],f(t)=[ ] f(t)=Pe^rt

In a petri dish there are initially 2520 bacteria After 2 hours there are 5040 bacteria.How many bacteria will there be after 12 hours? Let's begin by defining our variables.What is P? P=[?],t=[ ],f(t)=[ ] f(t)=Pe^rt
In a petri dish there are initially 2520 bacteria After 2 hours there are 5040 bacteria.How many bacteria will there be after 12 hours? Let's begin by defining our variables.What is P? P=[?],t=[ ],f(t)=[ ] f(t)=Pe^rt

Solution
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Answer

161280 bacteria after 12 hours. Explanation 1. Define Initial Variables P = 2520, initial number of bacteria. t is time in hours. f(t) is the number of bacteria at time t. 2. Calculate Growth Rate Use f(t) = Pe^{rt}. At t = 2, f(2) = 5040. Solve for r: 5040 = 2520e^{2r} \implies e^{2r} = 2 \implies 2r = \ln(2) \implies r = \frac{\ln(2)}{2}. 3. Predict Bacteria After 12 Hours Use f(t) = Pe^{rt} with t = 12: f(12) = 2520e^{\frac{\ln(2)}{2} \times 12} = 2520e^{6\ln(2)} = 2520 \cdot 2^6.

Explanation

1. Define Initial Variables<br /> $P = 2520$, initial number of bacteria. $t$ is time in hours. $f(t)$ is the number of bacteria at time $t$.<br /><br />2. Calculate Growth Rate<br /> Use $f(t) = Pe^{rt}$. At $t = 2$, $f(2) = 5040$. Solve for $r$: <br /> $5040 = 2520e^{2r} \implies e^{2r} = 2 \implies 2r = \ln(2) \implies r = \frac{\ln(2)}{2}$.<br /><br />3. Predict Bacteria After 12 Hours<br /> Use $f(t) = Pe^{rt}$ with $t = 12$: <br /> $f(12) = 2520e^{\frac{\ln(2)}{2} \times 12} = 2520e^{6\ln(2)} = 2520 \cdot 2^6$.
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