Suppose the population of a certain bacteria in a laboratory sample is 100.if it doubles in the population every 5hrs,what is the qrow rate?how many bacteria will there be in 3 days?

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Suppose the population of a certain bacteria in a laboratory sample is 100.if it doubles in the population every 5hrs,what is the qrow rate?how many bacteria will there be in 3 days?

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Answer 1 · 2018-06-26T12:47:42

Let $t$ $t$ be the time in hours, $P_{o}$ $P_o$ the initial population so $P = P_{o} e^{k t}$ $P=P_o e^{kt}$ is the population at $t$ $t$ . We're given $2 = e^{5 k}$ $2=e^{5k}$ so $k = \frac{1}{5} ln 2$ $k=1/5 ln 2$ . We're asked for $P$ $P$ at $t = 3 (24) = 72$ $t=3(24)=72$ which is

$P = 100 e^{(\frac{1}{5} ln 2) (72)} = 1638400 \cdot 2^{\frac{2}{5}} \approx 2161881$ $P= 100 e^{(1/5 ln 2)(72)} = 1638400 cdot 2^(2/5) approx 2161881$

Answer 2 · 2018-06-26T12:58:33

Growth rate is $0.138629$ $0.138629$ and bacteria population after
$72$ $72$ hours will be $2161882$ $2161882$

Explanation:

$B_{0} = 100, B_{5} = 200, B_{72} = ?, k = ?$ $B_0= 100 , B_5=200 , B_72= ? , k = ?$

$B_{t} = B_{0} \cdot e^{k t}; k, t$ $B_t = B_0*e^(kt); k , t$ are rate of growth and time in hours.

$B_{5} = B_{0} \cdot e^{k t} or 200 = 100 \cdot e^{k \cdot 5}$ $B_5 = B_0*e^(kt) or 200 =100 *e^(k*5)$ or

$e^{5 k} = \frac{200}{100} or e^{5 k} = 2$ $e^(5 k)= 200/100 or e^(5 k)= 2$ . Taking natural log on both

sides we get , $5 \cdot k \cdot ln (e) = ln (2) or k = \frac{ln (2)}{5}; [ln (e) = 1]$ $5*k*ln (e)= ln (2) or k = ln(2)/5 ; [ln (e)=1]$

$:.k~~0.138629$ . Growth rate is $0.138629$ and growth

equation is $B_t = B_0*e^(0.138629*t)$

Bacteria population after $72$ hours will be

$B_72 = 100*e^(0.138629*72)=2161882$ [Ans]

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Suppose the population of a certain bacteria in a laboratory sample is 100.if it doubles in the population every 5hrs,what is the qrow rate?how many bacteria will there be in 3 days?

2 Answers

Explanation:

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