# Assume that the number of bacteria follows an exponential growth model P(t)=Pe^{kt}? The count in the bacteria culture was 500 after 15 minutes and 2000 after 40 minutes. What was the initial size of the culture?

Sep 30, 2016

The initial size of the culture $P \left(0\right) = 125 \sqrt{4} \approx 165.$

#### Explanation:

$P \left(t\right) = P {e}^{k t} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(\star\right)$

$\text{When } t = 15 , P \left(15\right) = 500 \Rightarrow P {e}^{15 k} = 500. \ldots \ldots \ldots . \left(1\right)$

$w h e n t = 40 , P \left(40\right) = 2000 \Rightarrow P {e}^{40 k} = 2000. \ldots \ldots \ldots \ldots \left(2\right)$

$\left(2\right) \div \left(1\right) \Rightarrow \frac{P {e}^{40 k}}{P {e}^{15 k}} = \frac{2000}{500} \Rightarrow {e}^{25 k} = 4.$

$\Rightarrow {\left({e}^{25 k}\right)}^{\frac{1}{25}} = {4}^{\frac{1}{25}} \Rightarrow {e}^{k} = {4}^{\frac{1}{25}} \ldots \ldots \ldots \ldots \left(3\right)$

Now, rewriting $\left(1\right) \text{ as, "P(e^k)^15=500", and now, using } \left(3\right) ,$

we get, $P {\left({4}^{\frac{1}{25}}\right)}^{15} = 500 \Rightarrow P \left({4}^{\frac{3}{5}}\right) = 500$

Hence, $P = 500 \left({4}^{- \frac{4}{5}}\right) \ldots \ldots \ldots \ldots \ldots . . \left(4\right)$

(3),&, (4)", together with (star) "rArr P(t)=500(4^(-4/5))4^(t/25)

Finally, to get the initial size of the culture, P(0) , we have to take

$t = 0$ in this last eqn., so that,

$P \left(0\right) = 500 \left({4}^{- \frac{4}{5}}\right) {4}^{\frac{0}{25}} = 500 \left({4}^{- \frac{4}{5}}\right) = \frac{\left(500\right) \left(4\right) \left({4}^{- \frac{4}{5}}\right)}{4}$

$\therefore P \left(0\right) = \left(\frac{500}{4}\right) \sqrt{4} = 125 \sqrt{4} \approx \left(125\right) \left(1.3195\right) = 164.9375$

$\therefore P \left(0\right) \approx 165.$