# Math Help - Would I need to use logarithms here?

1. ## Would I need to use logarithms here?

If $
2^x4^y = 32
$

and $
\frac{3^x}{9^y}=3
$

then $
\frac{5^x}{125^y}= ?
$

This time the answer is $1$

I was trying to use the Logarithmic Properties to solve it, but I wasn't getting the correct answer.

How should I solve this?

$2^x4^y=32\Leftrightarrow 2^x2^{2y}=2^5\Leftrightarrow 2^{x+2y}=2^5\Leftrightarrow x+2y=5$ (1)
$\displaystyle \frac{3^x}{9^y}=3\Leftrightarrow \frac{3^x}{3^{2y}}=3\Leftrightarrow 3^{x-2y}=3^1\Leftrightarrow x-2y=1$ (2)
Solving the system formed by (1) and (2) you get $x=3, \ y=1$.
So $\displaystyle \frac{5^x}{125^y}=\frac{5^3}{125}=\frac{125}{125}= 1$