$a^{\log_g b} = b^{\log_g a}$?

Question

$g^{\log_g a} = a$, because it equals $a^{\log_g g}$.

Does this mean that $a^{\log_g b} = b^{\log_g a}$?

Note: thanks whoever edited it to proper markup

Answer 1

Yes, it is true that

$$a^{\log_g(b)}=b^{\log_g(a)}$$

You can see this by taking logs on both sides, and using $\log_g x^y=y \log_g x$.

Or, if you want a direct proof, let $x:= \log_g a$ and $y=\log_g b$. This means $$a=g^x \, \mbox{ and } b=g^y$$ You want to prove that $a^y=b^x$. This is easy:

$$a^y=(g^x)^y=g^{xy}=(g^{y})^x=b^x$$

Answer 2

Hint. One may recall the following results:

$$ \log (e^a)=a, \quad a \in \mathbb{R}, \tag1 $$ $$ a^b=e^{b \log a} \quad a>0.\tag2 $$

We have $$ a^{\log_g b}=e^{\large \frac{\log b}{\log g} \log a}=e^{\large \frac{(\log b)(\log a)}{\log g} } $$ and $$ b^{\log_g a}=e^{\large \frac{\log a}{\log g} \log b}=e^{\large \frac{(\log a)(\log b)}{\log g} } $$ then they are equal, since we always have

$$(\log b)\times (\log a)=(\log a)\times (\log b)$$

when everything is defined.

Answer 3

Using the formula $\displaystyle e^{\log_{e}(x)} = x.$

So we can write $\displaystyle a^{\log_{g}(b)} = g^{\log_{g}(a)^{\log_{g}(b)}} = g^{\log_{g}(b)\cdot \log_{g}(a)} = g^{\log_{g}(a)\cdot \log_{g}(b)} = g^{\log_{g}(b)^{\log_{g}(a)}} = b^{\log_{g}(a)}$

above we also used the formula $n\cdot \log_{x}(y) = \log_{x}(y)^n$