# Word problem - law of uninhibited growth

• Nov 24th 2008, 09:09 AM
rtwilton
A culture of bacteria obeys the law of uninhibited growth. N(t) = N0(e)^kt

If 500 bacteria are initially present and there are 80 after 1 hour, how many will be present after 5 hours? How long is it until there are 20,000 bacteria?

So, 800 = 500e^k and 800 = 1359.14^k

I cant get past this, how do I get the rate, divide each side by 1359.14?? Once I have the rate I can get the rest, correct?
• Nov 24th 2008, 09:19 AM
Henderson
Careful- you can't multiply 800 by e before you take care of your exponent.

Your best bet at this point:

$\displaystyle 800 = 500e^k$

$\displaystyle \frac{800}{500} = \frac{500e^k}{500}$

$\displaystyle 1.6 = e^k$

$\displaystyle ln(1.6) = k$
• Nov 24th 2008, 09:39 AM
rtwilton
Thank you!! So to answer the question how many bacteria will there be in 5 hours...

N(5) = 500e^ln(1.6)(5)

Hmmm this doesn't seem to work...

And for the other part of the question would it be

20,000 = 500e^ln(1.6)(5)

??