find the value $AB$, for which $\log_{A} B =\log_{B} A$ Suppose $A$ and $B$ are positive real numbers for which $\log_{A} B = \log_{B} A$ . If neither $A$ nor $B$ is $1$ and $A$ $\not\equiv B$, find the value of $AB$.
We can rewrite as
$A^{\log_{A} B} = B$.
$B^{\log_{B} A} = A$.
$(B^{\log_{B} A})^{A^{\log_{A} B}} =B$.
$B^{2 (\log_{B} A)} =B$.
$B^{2 (\frac{1}{2})} =B$.
I'm unable to deduce a correct solution with $\frac{1}{2}$.
I wanted to try a different approach by converting the powers to multiplication but can't figure out the base. i.e.
${\log_{A} B (\log_{?}A)} $.

To change from the base $b$ to the base $c$ in the logarithm below
$$\log_{b}a$$
we can use the following formula
$$\log_{b}a=\frac{\log_{c}a}{\log_{c}b}$$
So, to change from the base $B$ to base $A$:
$$\log_{B}A=\frac{\log_{A}A}{\log_{B}A}$$
and, since $\log_{A}A=1$, we get
$$\log_{B}A=\frac{1}{\log_{A}B}$$
and
$$\log_{B}A=\log_{A}B$$
then
$$(\log_{A}B)^{2}=1\Rightarrow\log_{A}B=\pm1$$
Therefore,
$$A=B$$
or
$$A=B^{-1}$$
