Exercise:
Solve in $\mathbb{R}$, ${{x}^{{{\log }_{2}}4x}}=8$
Solution: we know that $\log_{a(x)}{b(x)}$ is defined when $ b(x) >0 $ & $a(x)\ne 1 >0$
hence $\log_{2}{4x}$ is defined whenever $ x >0 $ now take logarithm both sides of base $x$
to get ${{\log }_{x}}{{x}^{{{\log }_{2}}4x}}={{\log }_{x}}8$ but $\log_{a}{b}=\frac{\log{a}}{\log{b}}$ so $\log_{2}{4x}=\frac{\log{4x}}{\log{2}}$
$\Rightarrow \frac{\log{4x}}{\log{2}}\times\log_{x}{x}=\frac{\log{8}}{\log{x}}$ but $\log_{x}{x}=1$ thus $\frac{\log{4x}}{\log{2}}=\frac{\log{8}}{\log{x}}$
but $\log{8}=\log{{2}^{3}}=3\log{2}$ hence $\frac{\log{4x}}{log{2}}= \frac{3\log{2}}{\log{x}}$
but $\log{\left(ab\right)}=\log(a)+log(b)$ thus $\log{(4x)}=\log{4}+\log{x}$
so $\frac{\log{4}+\log{x}}{\log{2}}=\frac{3\log{2}}{\log{x}}$
$\Rightarrow$ $ \log{x}\left(\log{4}+\log{x}\right)=3{\left(\log{2}\right)}^{2}=3\log^{2}{2}$
$\Rightarrow$ $2\log{2}\log{x}+\log^{2}{x}=3\log^{2}{2}$
Let $w=\log{x}$ $\Rightarrow$ $2\log{2}w+{w}^{2}-6\log{2}=0$ Adding both sides $\log^{2}{2}$ we get
${w}^{2}+2\log{2}w+\log^{2}{2}-\log^{2}{2}-3\log^{2}{2}=0$
$\Rightarrow$ ${\left(w+\log{2}\right)}^{2}=4\log^{2}{2}$ $\Leftrightarrow$ $w+\log{2}=\pm\sqrt{4\log^{2}{2}}=\pm2\log{2}$ thus $w=\log{2} \,\,or \,\,w=-3\log{2}$
but $w=\log{x} $ so $\log{x}=\log{2} \Rightarrow\, x=2 \,\, or \,\, \log{x}=\log{2}^{-3}\Leftrightarrow x={2}^{-3}=\frac{1}{8}$