A bacteria culture starts with 980 bacteria and grows at a rate
proportional to its size. After 6 hours there will be 5880 bacteria.
(a) Express the population after hours as a function of .
population:
(b) What will be the population after 4 hours?
© How long will it take for the population to reach 1490 ?

Question

A bacteria culture starts with 980 bacteria and grows at a rate
proportional to its size. After 6 hours there will be 5880 bacteria.
(a) Express the population after

hours as a function of

.
population:
(b) What will be the population after 4 hours?
© How long will it take for the population to reach 1490 ?

UpStudy AI · Accepted Answer

**(a)** We model the growth with the differential equation
$$
\frac{dP}{dt} = kP,
$$
whose solution is
$$
P(t)=P(0)e^{kt}.
$$
Given $$P(0)=980$$, we have
$$
P(t)=980e^{kt}.
$$
Since after 6 hours the population is 5880, we use
$$
980e^{6k}=5880.
$$
Dividing both sides by 980 yields
$$
e^{6k}=\frac{5880}{980}=6.
$$
Taking the natural logarithm of both sides gives
$$
6k=\ln 6,
$$
so that
$$
k=\frac{\ln6}{6}.
$$
Thus, the population as a function of time is
$$
P(t)=980e^{\frac{\ln6}{6}t}.
$$

**(b)** To find the population after 4 hours, substitute $$t=4$$ in the expression:
$$
P(4)=980e^{\frac{\ln6}{6}\cdot 4}=980e^{\frac{4\ln6}{6}}=980e^{\frac{2\ln6}{3}}.
$$
This can also be written as
$$
P(4)=980\cdot 6^{\frac{2}{3}}.
$$
Thus, the population after 4 hours is $$980\cdot 6^{\frac{2}{3}}$$.

**(c)** To determine the time $$t$$ when the population reaches 1490, we set
$$
1490=980e^{\frac{\ln6}{6}t}.
$$
Divide both sides by 980:
$$
\frac{1490}{980}=e^{\frac{\ln6}{6}t}.
$$
Taking the natural logarithm:
$$
\ln\left(\frac{1490}{980}\right)=\frac{\ln6}{6}t.
$$
Solving for $$t$$, we obtain
$$
t=\frac{6\ln\left(\frac{1490}{980}\right)}{\ln6}.
$$
**(a)** The population after $$ t $$ hours is $$ P(t) = 980 \times 6^{\frac{t}{6}} $$. **(b)** After 4 hours, the population is $$ 980 \times 6^{\frac{2}{3}} $$. **(c)** It takes approximately $$ \frac{6 \times \ln\left(\frac{1490}{980}\right)}{\ln6} $$ hours for the population to reach 1490.

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