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Exponential Functions Equation - WayOfTime - 2010-05-20
Today in grade 11 Functions we all got a question that I am unable to solve. Is it possible that you guys can help me solve it, showing how you got the answer?
In a forest there are twice as many grey moths as there are black ones. The population of the grey moth increases by 3% a month, whereas the black moths increase by 4% a month. How many months would it take for there to be twice as many black moths as grey moths?
I know it involves Log, but for this question I do not know how to use it. I know how to do standard log equations, but it doesn't seem that standard.
Exponential Functions Equation - shouri - 2010-05-20
2a e^(.03t) = a e^(.04t)
Find t that makes this true. You should have too much trouble solving this.
To see why I got this let G be the population of grey moths with G(0)=2a:
dG/dt =.03 G
^Change in grey moths per time
dG / G =.03 dt <~ I multiplied both sides by dt then divided both sides by G.
Integrate both sides:
ln(G) = .03 t +C
put a base of e on both of these.
G = e^(.03t +C)
G = e^(.03t) * e^© <~ this last part is just a constant, so we'll replace it with C
G = C * e^(.03t)
Use the fact that G(0) = 2a to get that C = 2a
G = 2a e^(.03t)
you do the same to get the population for B (the black moths)
Then just set them equal to each other and solve for t.
Exponential Functions Equation - Hazzy - 2010-05-20
Got all excited to work this out, and I see shouri has already done it. :[
Wolframming it can give you a value of t.
Exponential Functions Equation - WayOfTime - 2010-05-20
So we use Integrate instead of Log? Ok. Thanks!
Edit: It seems there is two ways to solve it? I'll just wait until tomorrow. The thanks still stands.
I have another question: Is this assuming that they start with the same value? As in equal amounts of both? Because the question says that it starts as twice as many grey moths than blacks.
Exponential Functions Equation - Noah - 2010-05-20
You could also use geometric series, there's really no need to integrate here.
Say we have n black moths, then the population of moths will be
![[Image: 396ugqe.png]](http://mathurl.com/396ugqe.png)
And the population of grey moths will be
![[Image: 27dfndv.png]](http://mathurl.com/27dfndv.png)
Solving for t:
![[Image: 32qvpnq.png]](http://mathurl.com/32qvpnq.png)
And the rest is pretty straightforward.
Noah
Exponential Functions Equation - shouri - 2010-05-20
Noah Wrote:You could also use geometric series, there's really no need to integrate here.
Say we have n black moths, then the population of moths will be
And the population of grey moths will be
Solving for t:
And the rest is pretty straightforward.
Noah
Actually Noah... this brings up an interesting point... Should we NOT count this as continuous growth? I thought we should because this is a population growth problem, thus I used the continuous growth model. But you used the non continuous exponential growth model that is usually used for things like interest which is applied only once in a given period:
![[Image: 3c61f664e4b9ae0ea85f89dff6b52548.png]](http://upload.wikimedia.org/math/3/c/6/3c61f664e4b9ae0ea85f89dff6b52548.png)
(Granted, you used geometric series... but that gives you the non continuous result)
WayOfTime Wrote:So we use Integrate instead of Log? Ok. Thanks!
Edit: It seems there is two ways to solve it? I'll just wait until tomorrow. The thanks still stands.
I have another question: Is this assuming that they start with the same value? As in equal amounts of both? Because the question says that it starts as twice as many grey moths than blacks.
And we assumed that they started with the values you told us about.
That's why we have 2a and a in the equations (in my case), and 2n and n in Noah's equation.
Also... when you solve for t... you get t in terms of a natural log. So natural logs WILL be involved in the answer.
Exponential Functions Equation - Russt - 2010-05-20
Shouri, whether or not it is continuous growth, the problem states that the population of black moth increases by 4% each month. That means that the ratio of the population in one month to the previous month should be 1.04.
If they're just learning exponential functions and not the calculus involving them I would think that they're looking for the 1.04^t approach.
Exponential Functions Equation - shouri - 2010-05-20
They didn't state whether they were learning them the calculus way or not 0.o
I first learned about them the calculus way now that I think about it.
Go with Noah's method... based off of what russt said.
Exponential Functions Equation - Noah - 2010-05-21
shouri Wrote:Actually Noah... this brings up an interesting point... Should we NOT count this as continuous growth? I thought we should because this is a population growth problem, thus I used the continuous growth model. But you used the non continuous exponential growth model that is usually used for things like interest which is applied only once in a given period:
(Granted, you used geometric series... but that gives you the non continuous result)
Short answer is already answered by Russt.
Long answer is that we should never, if a question is asked like this, integrate like that. This is actually not continuous growth, and we have to treat it as such. So, in fact, doing integration actually gives you the wrong answer (url). That's because, in your case, you don't think of it as growth over one month but rather that the increase by a percentage for every dt. Δy for any given t to t+1 should thus be 1.04/1.03, but yours grow faster because it includes the area between the diff to calculate the growth: 1 and 2.
Geometric series works perfectly well for continuous series as well, by the way.
Noah