Why does $(2^{\log_3n}=n^{\log_32})$?

Question

I'm trying to figure out which exponents/logarithms proprieties are used when we say that $2^{\log_3n}$ equals $n^{\log_32}$ just like Wolfram Alpha suggests in the alternative representation section, here.

Any ideas ? Thanks !

Answer 1

Well $2 = 3^{\log_3(2)}$ (this is part of the definition of log) and $n = 3^{\log_3(n)}$.

Thus:

$$2^{\log_3(n)} = \left( 3^{\log_3(2)} \right)^{\log_3(n)} = 3^{\log_3(n) \log_3(2)}$$

and

$$ n^{\log_3(2)} = \left( 3^{\log_3(n)} \right)^{\log_3(2)} = 3^{\log_3(n) \log_3(2)}$$.

Here we have used the property of exponents $(a^b)^c=a^{bc}$

Answer 2

$$2^{\log_3n}=n^{\log_32}$$ take the $\log_3$ for both sides $$\log_3n\log_32=\log_32\log_3n$$ Note that the sides are equal