# Exponential growth equation - C4Watch

#1
I've been trying to work out a question on this topic but haven't been able to get started properly. Could anyone tell me how to approach this?

The number, N, of bacteria increases at a rate proportional to the number present at any instant. After timing for 3 hours, the number is 5000 and after 5 hours, the number is 10 000. Find the initial number of bacteria.

I know the equation is

and that

, and

But I can't work out how to get values for and k.

It also seems clear that so

But beyond that I get nowhere except going round in circles saying and

It seems like the kind of thing that could be solved simultaneously, but I'm not sure and don't really know how do do it..!
0
quote
5 years ago
#2
(Original post by jonnburton)
I've been trying to work out a question on this topic but haven't been able to get started properly. Could anyone tell me how to approach this?

The number, N, of bacteria increases at a rate proportional to the number present at any instant. After timing for 3 hours, the number is 5000 and after 5 hours, the number is 10 000. Find the initial number of bacteria.

I know the equation is

and that

, and

But I can't work out how to get values for and k.

It also seems clear that so

But beyond that I get nowhere except going round in circles saying and

It seems like the kind of thing that could be solved simultaneously, but I'm not sure and don't really know how do do it..!
From your second line can you not say

my latex is awful, sorry.
0
quote
5 years ago
#3
5000=N0e^3k and 10000=N0e^5k

Equating and rearranging for N0

5000/e^3k = 10000/e^5k

Rearranging gives

e^2k = 2, now you can solve this for k.

Then using a given time t, a given number N and your value of k you can find N0 at these parameters.

You can also do the problem to simultaneously solve in terms of k first but I didnt want to try that in my head.
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quote
#4
Ah, I completely missed that! I'll try it... Thanks JWX!

(Original post by J_W-x)
From your second line can you not say

my latex is awful, sorry.
0
quote
5 years ago
#5
(Original post by jonnburton)
Ah, I completely missed that! I'll try it... Thanks JWX!
It's alright , sometimes it's good to be lazy, look for the easiest way of solving it possible without getting into messy ln business is what I always do first
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