I am reviewing logarithms and ran into this relationship $n^{log_a b} = b^{log_a n}$
Could use some help understanding it
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1It follows from $\log_a(b) \log n = \log_a(n) \log b$. Since $\log_a(b) = \frac{\log b}{\log a}$ and $\log_a(n) = \frac{\log n}{\log a}$ it is true. – egglog Jun 20 '22 at 17:08
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3Does this answer your question? Can we prove $a^{\log_bn} = n^{\log_ba}$? – Jun 20 '22 at 17:12
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Thank you https://math.stackexchange.com/users/983440/yooo for referencing a different answer. Searching by "n^{log_a b} = b^{log_a n} " led me to https://math.stackexchange.com/a/4199642/1068720 which gave me exactly what I was looking for.
The answer started with concepts and ended by explaining each operation.
This is the answer copied from https://math.stackexchange.com/users/544150/mike's post so people can see how good it is:
Are you familiar with the following: for any $c,d>1$: $d = c^{\log_c d}$. This follows from the definition of logarithm. I trust you also know the equation $e^{fg} = (e^f)^g$, for all positive e,f,g$
Thus: $$a^{\log_b n} = (b^{\log_b a})^{\log_b n}$$ $$= b^{\log_b a \log_b n} = (b^{\log_b n})^{\log_b a}$$ $$= n^{\log_b a}.$$
The first and last "=" following from the fact that $d = c^{\log_c d}$ for any $c,d>1$, and the remaining "=" following from the fact that the equation $e^{fg} = (e^f)^g$ holds for all positive ,,.
Ed Ames
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I would just note that $n = a^{\log_a n}$ and $b=a^{\log_a b}$.
Then you have
$$n^{\log_a b} =( a^{\log_a n})^{\log_a b}.$$
While
$$b^{\log_a n} = (a^{\log_a b})^{\log_a n}.$$
By the rules of exponents, the two right sides are equal.
