# Logirithms-- Doubling problem

• May 11th 2013, 01:54 AM
Gurp925
Logirithms-- Doubling problem
If some one could answer this it would be great.

A population of bacteria is quadrupling in a glass of milk on the counter every half hour.

A)Write a function for this growth P(t), where "t" is in hours
B)Write a function for this growth P(t), where "t" is in minutes
c)If the current bacteria population is 250000, how many bacteria were there 58 minutes ago?

Thanks!
• May 11th 2013, 03:23 AM
Plato
Re: Logirithms-- Doubling problem
Quote:

Originally Posted by Gurp925
If some one could answer this it would be great.
A population of bacteria is quadrupling in a glass of milk on the counter every half hour.

A)Write a function for this growth P(t), where "t" is in hours
B)Write a function for this growth P(t), where "t" is in minutes
c)If the current bacteria population is 250000, how many bacteria were there 58 minutes ago?

Surely, you are not asking us to do your work for you. Are you?

So that means you have tried something?
Post it so we can see where you are.
• May 11th 2013, 03:44 AM
Trefoil2727
Re: Logirithms-- Doubling problem
hello
I'm just trying..don't know whether right or not..
(a) P(t)=4^(2t)
(b) P(t)=4^(t/30)
(c) 4^(t/30)=250000
4^(t/30 -58/30)=250000(4^(-58/30))
=17137.89
• May 11th 2013, 03:59 AM
HallsofIvy
Re: Logirithms-- Doubling problem
Quote:

Originally Posted by Trefoil2727
hello
I'm just trying..don't know whether right or not..
(a) P(t)=4^(2t)
(b) P(t)=4^(t/30)

A good start but P(0)= 4^0= 1 and you were not told that. Put in a constant to represent the initial number of bacteria.

Quote:

(c) 4^(t/30)=250000
No, the problem said "the current bacteria is 250000". That is not true for any time, t. There are two ways to do this:
1) Take t= 0 as "current", find the constant so that P(0)= 250000, and then find P(-58).
2) Take t= 58 as "current", find the constant so that P(58)= 250000, and then find P(0).

Quote:

4^(t/30 -58/30)=250000(4^(-58/30))
=17137.89
• May 11th 2013, 01:27 PM
Gurp925
Re: Logirithms-- Doubling problem
This is what I got

A)P(t)=A(4^t.5)
B)P(t)=A(4^t/30)
• May 11th 2013, 01:40 PM
Gurp925
Re: Logirithms-- Doubling problem
So for C I solve For P(0) for the constant, then when i have the constant I can solve for P(-58)? But when solving for P(0)= 250000 I get P(0)=250000=A(240^0/30) It turns to the same answer of course so what am I doing wrong?

This is what I set up P(-58)=250000(4^-58/30)=17137.89 Seems as though it can be right seeing as it is less then 250000 lol.
• May 11th 2013, 01:41 PM
Gurp925
Re: Logirithms-- Doubling problem
Quote:

Originally Posted by Plato
Surely, you are not asking us to do your work for you. Are you?

So that means you have tried something?
Post it so we can see where you are.

Of course not my mistake for not posting my work I apologise, I have posted my work above.