Question

The initial population of a town is 3600, and it grows with a doubling time of 10 years. What will the population be in 8 years?

Answer 1

You should consider multiplying 3600 times 2^(something), i.e. \(\large 3600(2)^{\text{something}}\)

Now, you should consider \(x\) to be your time variable (or you can use \(t\) if you want). And you want the population to double, such that when you plug in x=10, you end up with the doubled population, i.e. \(3600(2)=7200\).

So in the exponent, when you plug in x = 10, you basically want your 2 to have an exponent of 1 (since \(\large 2^1=2\) ). So you need to multiply something by 10 to get 1 in the exponent, i.e.:
\[ y\times10=1\], then \(y = 1/10\), so this 1/10 is the factor you need to multiply your time variable x with.

Thus, you obtain:
\[ \large 3600(2)^{\frac{1}{10}x}\]

To see how this works: let's use x=10. Again, since the population must double in 10 years, we expect that you get 3600 x 2 = 7200 when you plug in x =10:
\[ \large 3600(2)^{\frac{1}{10}(10)}=3600(2)^{\frac{10}{10}}=3600(2)=7200\]

So to get the population after 8 years, just plug in x = 8.