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This is probably a very obvious question, but here goes...

An answer in my textbook claims that

$$3^{\log n} = n^{\log 3}$$

and that

$$4n^2 (3/4)^{\log n} = 4n^{\log 3}$$

Why, using more basic laws, is this the case?

(Unfortunately Google confuses this question with changing bases, exponentiation being the inverse of log (which is of course related), and similar matters.)

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    $\begingroup$ $3^{\log n} = e^{\log n\cdot \log 3}$, and $n^{\log 3} = e^{\log 3\cdot \log n}$. $\endgroup$ – Daniel Fischer Oct 10 '13 at 19:26
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Hint : $a=b$ implies $\log a=\log b$ and $\log a^b=b\log a$

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Recall that $3 = b^{\log_b 3}$.

Therefore $3^{\log_b n} = \Big(b^{\log_b 3}\Big)^{\log_b n} = b^{(\log_b 3)(\log_b n)} = \Big(b^{\log_b n}\Big)^{\log_b 3} = n^{\log_b 3}$.

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Change basis:

$$3^{\log n}=3^{\frac{\log_3n}{\log_3e}}=\left(3^{\log_3n}\right)^{\frac{1}{\log_3e}}=n^{1/\log_3e}=n^{\log 3}$$

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