use-the-formula-P-Ie-kt-where-P-is-resulting-population-I-is-the-initial-population-and-t-is-measured-in-hours-A-bacterial-culture-has-an-initial-population-of-10-000-If-its-declines-to-5000-in

Question Number 116976 by bobhans last updated on 08/Oct/20

use the formula P = {Ie}^{kt}, where P is resulting

population, I is the initial population and t is

measured in hours . A bacterial culture

has an initial population of 10, 000 . If

its declines to 5000 in 6 hours, what

will it be at the end of 8 hours ?

(a) 1985 (b) 3969 (c) 2500 (d) 4353

Answered by bobhans last updated on 08/Oct/20

\Rightarrow 5000 = 10, {000 e}^{6 k}

\Rightarrow \frac{5000}{10, 000} = e^{6 k} \Rightarrow \frac{1}{2} = e^{6 k}; 6 k = - \ln 2

k = - \frac{\ln 2}{6} = - 0.115525

we want compute P = 10, {000 e}^{8 k}

P = 10, {000 e}^{- 8 \times 0.115525} = 3969

Answered by Rio Michael last updated on 08/Oct/20

i will say P (t) = P_{0} e^{k t} just because am

use to writing it this way .

so P_{0} = 10, 000 bac, P (6) = 5000 bac

\Rightarrow 5000 = 10, 000 e^{6 k}

\Rightarrow \frac{1}{2} = e^{6 k}

\ln (\frac{1}{2}) = 6 k or k = - \frac{\ln 2}{6}

hence, P (t) = 10000 e^{- \frac{\ln 2}{6} t}

when t = 8, P (8) = 10, 000 e^{- \frac{4 \ln 2}{3}} = 3969

answer = (b)