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?

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Answer 1 · 2017-11-05T12:52:03

Bacteria population after $60$ $60$ minutes is $6000$ $6000$

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

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

next $30$ $30$ minutes . i.e in $60$ $60$ 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 $(p(30))/(p(0))$

$=(1500e^(30k))/(1500*e^0)=3000/1500 or e^(30k)=2$

Taking natural log on both sides we get

$30k ln e =ln2 or k= ln2/30=0.023105 [lne=1]$

Bacteria population after $t=60$ minutes is

$p(60)= 1500 e^(0.23105*60)= 6000$ [Ans]