A species of bacteria has a population of 250 at 6 a.m. It doubles every 8 hours. Determine the function that models the growth of the population. Determine the population at 9 p.m.?

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Answer 1 · 2018-05-28T10:50:41

6 a.m. to 2 p.m. population reaches 500.

Your equation is

(it is exponential function)

$P = P_{0} \times e^{k t}$ $P = P_0timese^(kt)$

$P_{=}$ $P_=$ is the initial population

t is time which is 15 hours.

First, You can compute k (growth rate)

$500 = 250 \times e^{k \times 8}$ $500=250timese^(ktimes8)$

$2 = e^{k \times 8}$ $2 = e^(ktimes8)$

$ln 2 = k \times 8$ $ln2 = ktimes8$

$k = 0.0866$ $k= 0.0866$ ${(h o u r)}^{- 1}$ $(hour)^-1$

Now you can calculate the final population

$P = 250 \times e^{0.0866 \times 15}$ $P=250timese^(0.0866times15)$

$P = 917$ $P=917$

which is greater than 500.