At the start of an experiment, there are 100 bacteria. If the bacteria follow an exponential growth pattern with rate k = 0.02, what will be the population after 5 hours? How long will it take for the population to double?

1 Answer
Jan 24, 2017

I tried this:

Explanation:

I would write the growth function as:
#y=100e^(kt)#
So that at #t=0# we get:
#y=100e^(0.02*0)=100# bacteria.
At #t=5# hours we will get:
#y=100e^(5*0.02)=110# bacteria.
To double the initial number we need to solve:
#200=100e^(0.02t)#
#e^(0.02t)=2#
Apply natural logs to both sides:
#ln(e^(0.02t))=ln(2)#
#0.02t=ln(2)#
#t=(ln(2))/0.02=34.65~~34.6#
#34# hours and 36 minutes approx.