# Math Help - Economic Development - Logarithms problem

1. ## Economic Development - Logarithms problem

Hi, I'm having a lot of difficulty with this problem because I'm terrible with logs.

Consider two countries, Chairland and Couchia. In year 0, Chairland starts out with a per-capita
income of $100, and this grows at a rate of 2% per year. Couchia starts out with a per-capita income of$1000, and this grows at a rate of 1% per year.
In how many years will Chairland’s per-capita income catch up with Couchia’s per-capita income?
(decimals are OK).

So my setup was:
x = years
100 * 1.02^x = 1000 * 1.01^x
= 1.02^x = 10 * 1.01^x

I have no idea where to go from here. Could anyone show me what I should do next?

2. Originally Posted by JoeZYe
So my setup was:
x = years
100 * 1.02^x = 1000 * 1.01^x
= 1.02^x = 10 * 1.01^x
$=x \ln{1.02} = \ln {10}+x \ln {1.01}$
so; $x = \frac{\ln{10}}{\ln{1.02}-\ln {1.01}}=233.71$

Some important log rules;
$\ln{xy}=\ln{x}+\ln{y}$
$\ln{\frac{x}{y}}=\ln{x}-\ln{y}$
$\ln{x^{n}}=n\ln{x}$

So you could have said;
$\left( \frac{1.02}{1.01} \right)^{x}=10$ and then taken logs and gotten the same answer...