Simplifying $\log_{\frac{1}{3}}(3^{2x})$ So I have the following problem: $\log_{\frac{1}{3}}(3^{2x})$ How do I solve this? Somewhere I stumbled onto the solution where they find a common exponent and base so they cancel, then you'll have your answer from what's left. However that was not very intuitive for me, although clever. I was trying "my" method where I translate it to a algebraic problem in terms of an exponentials (not sure about the nomenclature). E.g: $\left(\frac{1}{3}\right)^x = 3^{2x}$ But this gives me the wrong answer. Could somebody corroborate if this is a legitmate approach or not? Thank you in advance.

You have a good idea. But you need to define variables properly. Let $$ y=\log_{\frac{1}{3}}(3^{2x}) $$ Then $$ 3^{-y}=\left(\frac{1}{3}\right)^y = 3^{2x}. $$ The exponents in the previous equation must be the same (since the exponential function is injective). Thus $$ y=-2x. $$