Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
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 $x,y \in \mathbb{R}$.

share|cite|improve this answer
(+1) I've edited the layout of your answer. See what you think. (Love the profile pic, by the way.) – Cameron Buie Jul 30 '13 at 0:49
@CameronBuie: The layout you've provided is fine. (And thanks) :) – Sujaan Kunalan Jul 30 '13 at 1:06

Make base the same and equate exponents HINT : $ 3^2 = 9 $ (so make both LHS nad RHS base as 3 )

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.