Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A city doubles its population in 25 years. If it is growing exponentially, when will it triple its population?

The above is a question in my maths textbook in the topic Exponential Growth & Decay.

I'm a bit confused as to how I should approach this question.

We have been taught to use the formula: $$Q=Ae^{kt}$$ Where $Q$ is the quantity, $A$ is the initial quantity, $k$ is the growth/decay constant and $t$ is the time.

In reference to the question, I don't think I need $A$ so here is the equation I ended up with: $$2Q=e^{25k}$$


I found out that $$k=\frac{\ln2}{25}$$ I then let $Q=3A$ and the following is my working: $$3A=Ae^{25\frac{\ln2}{25}t}$$ $$3A=Ae^{\ln2t}$$ $$3=e^{\ln2t}$$ $$3=2^{t}$$ $$\ln{3}=t\ln{2}$$ $$t=\frac{\ln{3}}{\ln{2}}$$ $$t=1.6$$

I can't figure out what is wrong in my working out.

The provided answer is: 39.6 years

share|cite|improve this question
You do need $A$; after 25 years, the population is twice $A$, not twice $Q$. And you want to know when the population will be three times $A$. – Gerry Myerson Jun 25 '12 at 12:58
Exactly @Gerry well put, so what happens to the formulae when $Q=2A$ and then $Q=3A$? – Autolatry Jun 25 '12 at 13:03
You were careless when you let $Q=3A$. What happened to the $t$ in $Q=Ae^{kt}$? – Gerry Myerson Jun 26 '12 at 9:45
@GerryMyerson Right, included the $t$ this time but I am still stuck on what I am doing wrong. – Jallah Jun 26 '12 at 9:59
There's still a mistake in that very first line after "let $Q=3A$." You've got $Q=Ae^{kt}$; $k=(1/25)\log2$; $Q=3A$. Put them together carefully and you'll get to the right answer. – Gerry Myerson Jun 26 '12 at 10:04
up vote 1 down vote accepted

Start with $Q = Ae^{kt}$. If the doubling time is 25 years, this translates to $$2A = Ae^{25k}.$$ You should be able to solve for $k$ and make a go of it now.

share|cite|improve this answer
I've found that k is $\frac{\ln 2}{25}$. I'm not sure what to do now that I have k. – Jallah Jun 25 '12 at 13:15
juleszero, did you read my comment? Doesn't it suggest what to do now that you have $k$? – Gerry Myerson Jun 25 '12 at 13:28
@GerryMyerson Yes, I did read your comment but I am still getting the wrong answer hence why I am asking for further help. – Jallah Jun 26 '12 at 3:27
Show us what you've done; tell us what you think the right answer is, and why you think it's the right answer. When we've seen all of that, it will be much easier to tailor our responses to your situation. – Gerry Myerson Jun 26 '12 at 6:24
@GerryMyerson Thanks Gerry, I've updated my question to show my working. – Jallah Jun 26 '12 at 9:41

Hint: once you restore the A, you can divide the two equations to eliminate A and Q. That will allow you to evaluate k.

share|cite|improve this answer

Bro you were so close, you used ln a bit incorrectly though

Everything up to k = ln(2)/25 is correct

You just needed to do: 3A = Ae^kt (as the population is tripled)

Giving you 3 = e^kt, then take the ln of both sides to eliminate e

ln(3) = kt, and using your answer of k you find t

ln(3)/k = t or t = ln(3)/k

and therefore t = 39.6

share|cite|improve this answer
This is an old question which already has an accepted answer. You have provided no new insight. Please refrain from answering old questions which already have an answer, unless you are contributing something new – Shailesh Sep 26 '15 at 2:24

See what the constant of proportionality $k$ is equal to.

That is required time for doubling, tripling, quadrapuling ...

$$ k =\dfrac{ln\, 2}{ 25} =\dfrac{ln \, 3}{?} =\dfrac{ln\, 4}{?} =\dfrac{ln \,5}{?} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.