# Solve for $x$ without using logs: $3^{2x-1} = 9^{-x}$

Solve for $x$ without using logs: $3^{2x-1} = 9^{-x}$

Any clues on how to start this problem would be welcomed.

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Maybe use $9 = 3^2$? –  daniel Jul 30 '13 at 0:40

\begin{align}3^{2x-1}=9^{-x} &\implies3^{2x-1}=3^{2(-x)}\\ &\implies 2x-1=-2x\\ &\implies4x-1=0\\ &\implies4x=1\\ &\implies x=\dfrac{1}{4}\end{align}
The important thing is to recognize that $9=3^2$ and recall that if you have $a^x=a^y\implies x=y,$ for $x,y \in \mathbb{R}$.
Make base the same and equate exponents HINT : $3^2 = 9$ (so make both LHS nad RHS base as 3 )