Question

the count in a bacteria culture was 600 after 15 minutes and 1600 after 30 minutes. assuming the count grows exponentially ( $f(t)=ce^{kt}$ ),
find the initial size of the culture: (round to the nearest whole number.)
find the growth rate: $k =$ (round to the nearest thousandth.)
find the doubling period: $t_d=$ minutes. (round to the nearest tenth.)
find the population after 85 minutes: $p =$ (round to the nearest whole number.)
the population will reach 13000 at $t=$ minutes. (round to the nearest tenth.)

Explanation:

Step1: Set up equations

We know that $f(t)=ce^{kt}$. When $t = 15$, $f(15)=ce^{15k}=600$; when $t = 30$, $f(30)=ce^{30k}=1600$. Divide the second - equation by the first equation: $\frac{ce^{30k}}{ce^{15k}}=\frac{1600}{600}$. Using the property of exponents $\frac{e^{a}}{e^{b}}=e^{a - b}$, we get $e^{15k}=\frac{8}{3}$.

Step2: Solve for $k$

Take the natural logarithm of both sides of $e^{15k}=\frac{8}{3}$: $15k=\ln(\frac{8}{3})$. Then $k=\frac{\ln(\frac{8}{3})}{15}\approx\frac{\ln(8)-\ln(3)}{15}\approx\frac{2.0794 - 1.0986}{15}\approx0.0654$.

Step3: Solve for $c$

Substitute $k\approx0.0654$ and $t = 15$, $f(15)=600$ into $f(t)=ce^{kt}$. So $600 = ce^{15\times0.0654}$, and $c=\frac{600}{e^{15\times0.0654}}=\frac{600}{e^{0.981}}\approx\frac{600}{2.666}\approx225$.

Step4: Solve for the doubling - time $t_d$

We know that for exponential growth, when the population doubles, $f(t)=2c$. Substitute into $f(t)=ce^{kt}$: $2c=ce^{kt_d}$. Cancel out $c$ and get $2 = e^{kt_d}$. Take the natural logarithm of both sides: $\ln(2)=kt_d$. Then $t_d=\frac{\ln(2)}{k}\approx\frac{0.693}{0.0654}\approx10.6$.

Step5: Solve for the population after 85 minutes

Substitute $c\approx225$, $k\approx0.0654$, and $t = 85$ into $f(t)=ce^{kt}$: $P = 225e^{0.0654\times85}=225e^{5.559}\approx225\times259.07\approx58291$.

Step6: Solve for $t$ when $P = 13000$

Substitute $P = 13000$, $c\approx225$, $k\approx0.0654$ into $P = ce^{kt}$: $13000 = 225e^{0.0654t}$. Then $e^{0.0654t}=\frac{13000}{225}\approx57.7778$. Take the natural logarithm of both sides: $0.0654t=\ln(57.7778)\approx4.051$. So $t=\frac{\ln(57.7778)}{0.0654}\approx62.0$.

Answer:

Initial size of the culture: 225
Growth rate $k$: 0.065
Doubling - time $t_d$: 10.6
Population after 85 minutes $P$: 58291
$t$ when population reaches 13000: 62.0