# Thread: System of equation

1. ## System of equation

3^ log x = 4^log y
(4x) ^log 4 = (3y) ^log 3

Thanks!

2. ## Re: System of equation

Taking the logarithm of both sides of the first equation, (log(x))(log(3))= (log(y))(log(4)). Taking the logarithm of both sides of the second equation, (log(4)(log(4x)= log(4)(log(4)+ log(x))= (log(3))(log(3y))= (log(3))(log(3)+ log(y)) or (log(4))^2+ log(4)log(x)= (log(3))^2+ log(3)log(y).

From the first equation, $log(y)= \frac{log(3)}{log(4)}log(x)$. Putting that into the second equation
$(log(4))^2+ log(4)log(x)= (log(3))^2+ \frac{(log(3))^2}{log(4)}log(x)$

$\left((log(4)- \frac{(log(4))^2}{log(3)}\right)log(x)= (log(3))^2- (log(4))^2$
$\left(\frac{log(3)log(4)- (log(4))^2}{log(3)}\right)log(x)= (log(3))^2- (log(4))^2$
$log(x)= \frac{((log(3))^2- (log(4))^2)log(3)}{log(3)log(4)- (log(4))^2}$

Now use $log(y)= \frac{log(3)}{log(4)}log(x)$ to find $log(y)$ and take exponentials to find x and y.

You might be able to simplify that a bit using the facts that $(log(3))^2- (log(4))^2= (log(3)- log(4))(log(3)+ log(4))= log\left(\frac{3}{4}\right) log(12)$ and $log(3)log(4)- (log(4))^2= log(4)(log(3)- log(4))= log(4)log\left(\frac{3}{4}\right)$.