$\text { Suppose a population of bacteria is changing at a rate of } \frac { d p } { d t } = \frac { 1,000 } { 1 + 0.5 t } \text {, where } t \text { is }$ the time in days. The initial population (when $t = 0$ ) is 500 . Write an equation that gives the population at any time $t$.
A) $p ( t ) = 2,000 \ln | 1 + 0.5 t | + 250$
B) $p ( t ) = \ln | 1 + 0.5 t | + 500$
C) $p ( t ) = 1,000 \ln | 1 + 0.5 t | + 500$
D) $p ( t ) = 1,000 \ln | 1 + 0.5 t | + 250$
E) $p ( t ) = 2,000 \ln | 1 + 0.5 t | + 500$

Question

Quizplus · Accepted Answer

To find the population function $p(t)$, integrate the rate function $\frac{dp}{dt} = \frac{1000}{1 + 0.5t}$. The integral is $p(t) = 2000 \ln |1 + 0.5t| + C$. Given $p(0) = 500$, solve for $C$: $500 = 2000 \ln(1) + C$, so $C = 500$. Thus, $p(t) = 2000 \ln |1 + 0.5t| + 500$.

The Time in Days t=0t = 0t=0 ) Is 500 Write an Equation That Gives the Population at Any

The Time in Days $t = 0$ ) Is 500 Write an Equation That Gives the Population at Any