# A bacteria culture initially contains 1500 bacteria and doubles every half hour. The formula for the population is p(t)= 1500e^kt for some constant k, how do you find the size of the population after 60 minutes?

Nov 5, 2017

Bacteria population after $60$ minutes is $6000$

#### Explanation:

$1500$ bacteria doubles to $1500 \cdot 2 = 3000$ in $30$ minutes

$3000$ bacteria doubles to $3000 \cdot 2 = 6000$ in

next $30$ minutes . i.e in $60$ minutes.

General procedure for any time required:

p(t)= 1500 e^(kt) (1) ; :. p(0)= 1500*e^0=1500 (1) and

:p(30)= 1500e^(30k);(2) =3000 . Dividing equation (2) by

equation (1) we get $\frac{p \left(30\right)}{p \left(0\right)}$

$= \frac{1500 {e}^{30 k}}{1500 \cdot {e}^{0}} = \frac{3000}{1500} \mathmr{and} {e}^{30 k} = 2$

Taking natural log on both sides we get

$30 k \ln e = \ln 2 \mathmr{and} k = \ln \frac{2}{30} = 0.023105 \left[\ln e = 1\right]$

Bacteria population after $t = 60$ minutes is

$p \left(60\right) = 1500 {e}^{0.23105 \cdot 60} = 6000$ [Ans]