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I am struggling with what I think should be some a basic log problem:

Show that $3^{log_2n} = n^{log_23}$

I know that $3^{log_3n} = n$ and $log_2n = {log_3n}/{log_32}$

I was attempting something similar to:

$3^{{log_3n}/log_32} = 3^{log_3n - log_32}$ but then I got stuck. Am I on the right track by using the change of base and then subtracting?

EDIT: I'm not trying to exactly 'show that' the two expressions are equal. I am looking to figure out how to simplify the expression on the left to the one on the right

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In general, $$ n^{(\log x)} = (e^{\log n})^{(\log x)} = e^{(\log n)(\log x)} = e^{(\log x)(\log n)}= (e^{\log x})^{(\log n)} = x^{(\log n)} $$ where $\log = \log_b$ for fixed base $b$.

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  • $\begingroup$ Is there an actual name for this rule? After going through several sites containing rules of logarithms, I've not found this one being included. Having the rule's name would be helpful, if one exists. $\endgroup$
    – code_dredd
    Sep 26 '17 at 6:28
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    $\begingroup$ This answer is a bit miss leading as the OP is using base 2 not base e. The concept is correct though, $3 = 2^{log_23}$. The original power of 3 can be swapped with this new power since it is a product i.e. $3^{log_2n} = 2^{(log_23)(log_2n)} = 2^{(log_2n)(log_23)} = n^{log_23}$ $\endgroup$ Jul 25 '18 at 4:13
  • $\begingroup$ The log base e threw me off here. This is actually a very simple operation of taking the log of both sides. $\endgroup$ Dec 18 '20 at 17:13
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$$A=3^{\log_2n}\implies\log_2A=\log_2n\cdot\log_23$$

$$B=n^{\log_23}\implies\log_2B=\log_23\cdot\log_2n$$

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