word problem: Suppose bacteria culture doubles every....

[vintagelullabyx3]

Suppose that the population in a bacteria culture doubles every 20 minutes. At 110 minutes, there are 60,000 bacteria. What is the initial population? What is the population after 5 years?

I know that I need to make some sort of equation to get the 5 year thing, but I can't figure it out. I thought the initial populaltion was 1500 but it's not so now I'm lost.

 

[skeeter]

\(\displaystyle \L P = P_0 e^{kt}\)

at t = 20 minutes, the population doubles ...

\(\displaystyle \L 2P_0 = P_0 e^{20k}\)

\(\displaystyle \L 2 = e^{20k}\)

\(\displaystyle \L ln(2) = 20k\)

\(\displaystyle \L \frac{ln(2)}{20} = k\)

at t = 110 minutes, the population is 60000 ...

\(\displaystyle \L 60000 = P_0e^{110k}\)

\(\displaystyle \L \frac{60000}{e^{110k}} = P_0\)

you now have k and P₀ ... convert 5 years to seconds and use the first equation to find the unrealistically enormous population after that time.

 

[soroban]

Re: Stupid word problem.. please help..

Hello, vintagelullabyx3!

Suppose that the population in a bacteria culture doubles every 20 minutes.
At 110 minutes, there are 60,000 bacteria.
What is the initial population?
What is the population after 5 years?

The "doubling function" is: \(\displaystyle \,P \:=\_o\cdot2^n\)

where \(\displaystyle P_o\) is the initial population, and \(\displaystyle n\) is the number of 20-minute periods.


We are told: at 110 minutes \(\displaystyle (n\,=\,5.5),\;P\,=\,60,000\)

So we have: \(\displaystyle \,P_o\cdot2^{5.5}\,=\,60,000\;\;\Rightarrow\;\;\L P_o\,\approx\,1326\)


Are you sure they said "5 years"?

\(\displaystyle 5\text{ years} \,=\, 1825\text{ days} \,=\,43,800\text{ hours} \,=\,131,400\) twenty-minute periods

The population will be: \(\displaystyle \L\,P\:=\:1326\left(2^{131,400}\right)\)


FYI: This answer begins 29105... and has almost 40,000 digits.


Edit: Corrected my error . . . Thanks for the heads-up, skeeter!