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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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2 Answers 2



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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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