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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.

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    $\begingroup$ If I understand correctly, your approach involves letting the value of $\log _\frac{1}{3} (3^2x)$ be a particular value and working from there. You cannot set that particular value to be $x$, but you must set it to another variable. $\endgroup$ – Harry Alli Dec 19 '17 at 1:11
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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 $$ y=-2x. $$

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  • $\begingroup$ Ah, very helpful. Cheers! $\endgroup$ – oxodo Dec 19 '17 at 1:16
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Here is a minimal set of definitions you can always rely on :

  • $\log_a(x) = \dfrac{\ln(x)}{\ln(a)}$
  • $x^y=\exp(y\ln(x))$

$$\log_{\frac 13}(3^{2x})=\dfrac{\ln(3^{2x})}{\ln(\frac 13)}=\dfrac{\ln(\exp(2x\ln(3)))}{\ln(\frac 13)}=\dfrac{2x\ln(3)}{-\ln(3)}=-2x$$


I also used $\quad\ln(\exp(x))=x\quad$ and $\quad\ln(\frac 1x)=-\ln(x)\quad$ but I'm sure these do not pose problem.

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