# How do you find the size of the bacterial population after 100 minutes if a bacteria culture initially contains 1500 bacteria and doubles every half hour and the formula for the population is, p (t) = 1500e^(kt) for some constant, k ?

Mar 31, 2015

$p \left(100\right) = 15119$

#### Explanation:

Given $p \left(0\right) = 1500$, $p \left(30\right)$ will be

$3000 = 1500 \cdot {e}^{30 k}$

That is

$2 = {e}^{30 k}$.

Take log on both sides to get

$30 k = \ln 2 \implies k = \ln \frac{2}{30}$

Now

$p \left(100\right) = 1500 \cdot {e}^{100 k}$

$p \left(100\right) = 1500 \cdot {e}^{\frac{10}{3} \ln 2}$

p(100)= 1500 * e^(ln(2^(10/3))

$p \left(100\right) = 1500 \cdot {2}^{\frac{10}{3}}$

$p \left(100\right) = 1500 \cdot 10.079 = 15119$