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I saw this particular line slammed in a proof and it bothers me I can't understand why this is obvious and how would one justify this : $$ 7^{\log (n)} = n^{\log (7)} $$

Can anyone explain ?

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  • $\begingroup$ Hint: try writing $n$ as $e^{\log n}$ and then massaging the resulting expression a bit. $\endgroup$ – Dilip Sarwate Nov 12 '14 at 12:31
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$$7^{\ln n}=\exp (\ln 7 \ln n)=n^{\ln 7}.$$

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Take $\log$ of the LHS

\begin{equation} \begin{aligned} \log{(7^{\log{n}}}) =& (\log{n})(\log{7}) \\ =& \log{(n^{\log{7}}}) \end{aligned} \end{equation}

Then removing the log we took earlier we can clearly see $7^{\log{n}} = n^{\log{7}}.$

Hope this helps.

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  • $\begingroup$ I think you need to make a comment about the $\log$ function. For example, $(+1)^2=(-1)^2$ but you can't just remove the square to derive $+1=-1$... $\endgroup$ – JP McCarthy Nov 12 '14 at 17:12
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$$7^{\log n}=7^{\frac{\log_7(n)}{\log_7(10)}}=7^{\log_7(n^{1/\log_7(10)})}=n^{1/\log_7(10)}=n^{\log_7(7)/\log_7(10)}=n^{\log(7)}$$

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I think everyone is missing the point that the definition for $x^\alpha$ for irrational $\alpha$ is

$$x^\alpha =e^{\alpha\ln x}.$$

This is why the other answers make sense.

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  • $\begingroup$ I think actually that's exactly the property that the accepted answer used. Answers that took different approaches ended up more complicated, I think. $\endgroup$ – David K Nov 12 '14 at 13:34
  • $\begingroup$ I agree but it is a definition rather than a property. We only have $\exp(\ln 7\ln n)=n^{\ln 7}$ by definition of $n^\alpha$ for $\alpha$ irrational. $\endgroup$ – JP McCarthy Nov 12 '14 at 15:16
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    $\begingroup$ I was using the word "property" as a superset of "definition." One could also say this property holds by definition. $\endgroup$ – David K Nov 12 '14 at 16:58

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