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I have three populations

a <- 100 
b <- 100000
c <- 75000

Population a growths 10% per year while population b growths 1% per year and population c growths 2% per year.

How many years are necessary for the total population of a be the same of b and of c?

I have tried to solve this problem by using the compound interest formula (but I am not sure whether this is the right way to do it).

If so, how can I find t in this comparison so that it returns TRUE?

a * ((1 + 0.1) ^ (1 * t)) == b * ((1 + 0.01) ^ (1 * t))

Thanks,

1 Answers 1

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The population of A after $t $ years is $100 (1+.1)^t$.

B is $100,000 (1+.01)^t $

C is $75,000 (1+.02)^t$

You can find when any two of them are equal by setting 2 equal to each other and solving for $t $. But there's no reason to believe there will be a time all three are equal. There might be. But there's no reason to assume there is.

So $A =B $ when $100*1.1^t= 100000*1.01^t $ or when $A/B=1$ so $\frac {100*1.1^t}{100000*1.01^t}=1$

So $\frac {100}{100000}(\frac {1.1}{1.01}^t)=1$

$10^{-3}(\frac {1.1}{1.01})^t=1$

$ (\frac {1.1}{1.01})^t=10^3$

$t=\log_{\frac {1.1}{1.01}}10^3$

$=\frac {\log_{10}10^3}{\log_{10}\frac {1.1}{1.01}} =\frac 3 {\log_{10}\frac {1.1}{1.01}}$

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    Thanks, @fleablood. You are right. I should not assume that they will be equal in t years. Cheers, – 2017-02-20
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    Thanks for solving it @fleablood. I was not sure how to solve it for 't' anyways. Now you reminded me of my Math classes. Cheers, – 2017-02-20