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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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1 Answer 1

$$A=3^{\log_2n}\implies\log_2A=\log_2n\cdot\log_23$$

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

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