Prove $f^{\log_a g}=g^{\log_a f}$

Question

Prove $f^{\log_a g}=g^{\log_a f}$

$$ y=f^{\log_a g}\implies \log_fy=\log_ag\implies g=a^{\log_fy}\\ g^{\log_a f}=a^{{\log_a f}.{\log_f y}}=a^{{\log_af}.{\log_ag}}=a^{{\log_ag}.{\log_af}}=a^{\log_af^{\log_ag}}=f^{\log_ag} $$ Is it the right way to prove it ?

Answer 1

Hint:

$$x^y = a^{\log_a(x^y)}=a^{y\cdot \log_a x}$$

Answer 2

The function $y = \log_a x$ is one-to-one. Alternatively you could show that both sides have the same logarithm base $a$.

Answer 3

Let $y = \log_{a}(g)$ so that:

$$f^{\log_{a}(g)} = f^{y} = g^{\log_{g}(f^y)} = g^{y\log_{g}(f)} = g^{\log_{a}(g)\log_{g}(f)}$$

If we can show:

$$\log_{a}(g)\log_{g}(f) = \log_{a}(f)$$

we are done.

Observe:

$$\log_{a}(g)\log_{g}(f) = \log_{a}(g)\frac{\log_{a}(f)}{\log_{a}(g)} = \log_{a}(f)$$

Thus,

$$g^{\log_{a}(g)\log_{g}(f)} = g^{\log_{a}(f)}$$

(You can do this without introducing $y$, but I think it is easier to see with $y$)