Question

A certain strain of bacteria that is growing on your kitchen counter doubles every 15 minutes. Assuming that you start with only one bacterium, how many bacteria could be present at the end of 7 hours?
Use the equation A = A0ekt

Answer 1

there are 4 15 minute periods in each hour, times 7 hours is 28 doubling times
compute \(2^{28}\) it will be really really big

Answer 2

i was able to find that answer but is there anyway to use that formula..what would i substitute for each variable

Answer 3

You are given the following:
when t = 0, A = 1. Put this in the equation and solve for A0.
when t = 1/4 hour, A = 2. Put this in the equation and solve for k

Then put t = 7 and find A.

Answer 4

wow, way over my head..lol thanks for trying to help

Answer 5

yeah mine too
why would you need
\[A=A_0e^{kt}\] when you are actually told the doubling time? no need at all

Answer 6

im not too sure, thats just the way the question was stated...i guess anything to make it more difficult...lol

Answer 7

you can solve
\[e^{\frac{1}{4}k}=2\] if you like, in two steps
\[e^{\frac{1}{4}k}=2\\
\frac{1}{4}k=\ln(2)\\
k=4\ln(2)\]

Answer 8

then compute
\[e^{4\ln(2)\times 7}=e^{28\ln(2)}\] you will get the same answer

Answer 9

and that is because
\[28\ln(2)=\ln(2^{28})\] making
\[e^{28\ln(2)}=e^{\ln(2^{28})}=2^{28}\] so you are right back where you started from

Answer 10

Oh ok, cool, thanks a lot...that helps me a lot

Answer 11

really?
yw

Answer 12

yes, thanks a lot...It will have to do, if they dont like it, oh well.lol thanks again

Answer 13

yw again