Question

the count in a bacteria culture was 200 after 20 minutes and 1200 after 35 minutes. assuming the count grows exponentially. what was the initial size of the culture? find the doubling period. find the population after 80 minutes. when will the population reach 10000.

Explanation:

Step1: Set up exponential - growth formula

The general form of an exponential - growth model is $P(t)=P_0e^{kt}$, where $P(t)$ is the population at time $t$, $P_0$ is the initial population, and $k$ is the growth constant.
We know that $P(20) = 200$, so $200=P_0e^{20k}$, and $P(35)=1200$, so $1200 = P_0e^{35k}$.

Step2: Solve for $k$

Divide the second equation by the first equation: $\frac{1200}{200}=\frac{P_0e^{35k}}{P_0e^{20k}}$.
Simplify the left - hand side to get $6$, and the right - hand side using the property of exponents $\frac{e^{a}}{e^{b}}=e^{a - b}$ gives $e^{15k}$. So, $6 = e^{15k}$.
Take the natural logarithm of both sides: $\ln(6)=15k$. Then $k=\frac{\ln(6)}{15}\approx0.116$.

Step3: Solve for $P_0$

Substitute $k$ into the equation $200 = P_0e^{20k}$.
$200=P_0e^{20\times\frac{\ln(6)}{15}}=P_0e^{\frac{4}{3}\ln(6)}=P_06^{\frac{4}{3}}$.
$P_0=\frac{200}{6^{\frac{4}{3}}}\approx19.75\approx20$.

Step4: Find the doubling period $T$

Set $P(T) = 2P_0$. Using $P(t)=P_0e^{kt}$, we have $2P_0=P_0e^{kT}$.
Cancel out $P_0$ and get $2 = e^{kT}$. Take the natural logarithm of both sides: $\ln(2)=kT$.
Since $k=\frac{\ln(6)}{15}$, then $T=\frac{15\ln(2)}{\ln(6)}\approx6.8$ minutes.

Step5: Find the population after $t = 80$ minutes

$P(80)=P_0e^{80k}$. Substitute $P_0\approx20$ and $k=\frac{\ln(6)}{15}$ into the formula.
$P(80)=20e^{80\times\frac{\ln(6)}{15}}=20\times6^{\frac{16}{3}}\approx77760$.

Step6: Find the time $t$ when $P(t)=10000$

Set $10000 = P_0e^{kt}$. Substitute $P_0\approx20$ and $k=\frac{\ln(6)}{15}$ into the formula:
$10000 = 20e^{\frac{\ln(6)}{15}t}$.
$\frac{10000}{20}=e^{\frac{\ln(6)}{15}t}$, so $500 = e^{\frac{\ln(6)}{15}t}$.
Take the natural logarithm of both sides: $\ln(500)=\frac{\ln(6)}{15}t$.
$t=\frac{15\ln(500)}{\ln(6)}\approx62.1$ minutes.

Answer:

Initial size: $20$
Doubling period: $6.8$ minutes
Population after 80 minutes: $77760$
Time when population reaches 10000: $62.1$ minutes