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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: $2Q=e^{25k}$

Edit:

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

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    Exactly @Gerry well put, so what happens to the formulae when $Q=2A$ and then $Q=3A$?2012-06-25

4 Answers 4

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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.

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    @Gerry$M$yerson Thanks Gerry, I've updated my question to show my working.2012-06-26
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Hint: once you restore the A, you can divide the two equations to eliminate A and Q. That will allow you to evaluate k.

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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

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    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 new2015-09-26
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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}{?} $