78.1k views
0 votes
In a lab experiment, 2400 bacteria are placed in a petri dish. The conditions are such

that the number of bacteria is able to double every 19 hours. How long would it be, to
the nearest tenth of an hour, until there are 3300 bacteria present?

User Dtc
by
6.4k points

1 Answer

0 votes

Answer:

It takes 8.7 hours for there to be 3300 bacteria present.

Explanation:

Exponential equation for population growth:

The exponential equation for population growth is given by:


P(t) = P(0)(1+r)^t

In which P(0) is the initial population and r is the growth rate.

2400 bacteria are placed in a petri dish.

This means that
P(0) = 2400

So


P(t) = P(0)(1+r)^t


P(t) = 2400(1+r)^t

The conditions are such that the number of bacteria is able to double every 19 hours.

This means that
P(19) = 2P(0). We use this to find 1 + r.


P(t) = P(0)(1+r)^t


2P(0) = P(0)(1+r)^(19)


(1+r)^(19) = 2


\sqrt[19]{(1+r)^(19)} = \sqrt[19]{2}


1 + r = 2^{(1)/(19)}


1 + r = 1.03715504445

So


P(t) = 2400(1+r)^t


P(t) = 2400(1.03715504445)^t

How long would it be, to the nearest tenth of an hour, until there are 3300 bacteria present?

This is t for which
P(t) = 3300. So


P(t) = 2400(1.03715504445)^t


2400(1.03715504445)^t = 3300


(1.03715504445)^t = (3300)/(2400)


\log{(1.03715504445)^t} = \log{(33)/(24)}


t\log{1.03715504445} = \log{(33)/(24)}


t = \frac{\log{(33)/(24)}}{\log{1.03715504445}}


t = 8.7

It takes 8.7 hours for there to be 3300 bacteria present.

User Jeff Guillaume
by
6.3k points