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How to simplify this?

$\displaystyle\frac{n^{\log m}}{m^{\log n}}$

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Try rewriting each base. – Benji Jan 21 '11 at 1:35
possible duplicate of How $a^{\log_b x} = x^{\log_b a}$ ? – Bill Dubuque Jan 21 '11 at 2:38
up vote 3 down vote accepted

First note that $x = a^{\log_a x}$

So we find that $n^{\log m} = e^{\log (n^{\log m})} = e^{\log m \log n}$

Similarly, we find that $m^{\log n} = e^{\log (m^{\log n})} = e^{\log n \log m}$

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Hint: Apply $\log$ to the whole thing and use the quotient and powers rules for $\log$. You should get a very simple result, which you can exponentiate to find your answer.

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Note that $n^{\log m} = m^{\log n}$.

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I laughed at this. – Benji Jan 21 '11 at 1:36
Sorry I stated the wrong question. It should had been "How to prove that those expressions are equal" +1 – Wilbert Barrera Jan 21 '11 at 1:43
@Wilbert Barrera: To prove that, you can apply $\log$ to both sides, and use the fact that $\log$ is one-to-one. – Jonas Meyer Jan 21 '11 at 1:47

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