Question

Hw21-Obj-C6: Problem 9 Problem Value: 1 point(s). Problem Score: 0%. Attempts Remaining: 10 attempts. (1 point) The count in a bacteria culture was 700 after 20 minutes and 1900 after 35 minutes. Assume the bacteria population is growing exponentially What was the initial size of the culture? Σ bacteria Find the doubling period. minutes Find the population after 120 minutes. Σ bactoria When will the population reach 10000? Σ minutes Help Entering Answers Preview My Answers Submit Answers

Algebra

Answer 1

PROBLEM 9:

PART 1:

We know that, No.of bacteria at time t is given by,

$P_{(t)}=P_{o}.e^{rt}$

Where r is the growth constant and P_o is the initial size of culture.

We have,

$700=P_{o}.e^{r\times 20\;min}\rightarrow \left ( 1 \right )$

$1900=P_{o}.e^{r\times 35\;min}\rightarrow \left ( 2 \right )$

$\frac{(1)}{(2)}\Rightarrow \frac{700}{1900}=\frac{P_{o}e^{r\times 20}}{P_{o}e^{r\times 35}}$

$\therefore \frac{7}{19}=\frac{e^{r\times20}}{e^{r\times 35}}$

$\therefore \frac{7}{19}=e^{r(20-35)}$

Taking logarithm on both sides,

$\ln \left (\frac{7}{19} \right )=\ln\left (e^{r(20-35)} \right )$

$\Rightarrow \ln \left (\frac{7}{19} \right )=-15 r$

$\therefore r=\frac{-1}{-15}=\frac{1}{15}$

From (1) ,initial size is given by,

$700=P_{o}.e^{\frac{1}{15}\times 20\;min}$

$700=P_{o}.e^{1.333}$

$\therefore P_{o}=\frac{700}{e^{1.333}}$

$\mathbf{\therefore P_{o}=185\;bacteria}$

PART 2:

Doubling time t is given by,

$P_{(t)}=2P_{o}$

$\therefore P_{(t)}=2P_{o}=P_{o}.e^{rt}$

$\therefore e^{rt}=2$

Taking logarithm on both sides,

$\therefore \ln\left (e^{rt} \right )=\ln 2$

$\therefore rt=\ln 2$

$\therefore t=\frac{\ln 2}{r}=\ln 2\times 15$

$\mathbf{\therefore t=10.4\;min}$

PART 3:

Population after 120 min is given by,

$P_{(120)}=P_{o}.e^{r\times 120}=185\times .e^{1/15\times 120}$

$\Rightarrow P_{(120)}=185\times e^{8}$

$\mathbf{\therefore P_{(120)}=551477\;bacteria}$

PART 4:

$P_{t}=10000=P_{o}e^{rt}$

$e^{rt}=\frac{10000}{P_{o}}=\frac{10000}{185}=54.054$

Taking logarithm on both sides,

$\ln \left (e^{rt} \right )=\ln \left (54.054 \right )$

$rt=3.98998$

$\therefore t=\frac{3.98998}{r}=3.98998\times 15\;min$

$\mathbf{\therefore t=59.85\;min\approx 60 min}$