How is this identical transformation true $x^{1-\log(x)}=1\Longleftrightarrow \log x^{1-\log(x)}=\log1$?

Question

How is this identical transformation true $$x^{1-\log x}=1\Longleftrightarrow\log x^{1-\log x}=\log1\text{ ?}$$ I thought to put both sides on log: $$\log x^{1-\log x}=\log1,$$ but then I don't know if I can do this

Answer 1

$\require{cancel}$You can. You are just applying $\log: \Bbb R_{>0} \to \Bbb R$, which is a well-defined function, on the same element. In general: $${\rm stuff }\,1={\rm stuff}\,2 \implies f({\rm stuff}\,1) = f({\rm stuff}\,2)$$always that $f$ is a well-defined function and ${\rm stuff}\,1,{\rm stuff}\,2 \in {\rm dom}(f)$. This proves the $\implies$ part. Now we want to check that: $$\log x^{1-\log x} = \log 1 \implies x^{1-\log x} = 1.$$ It goes the same as above, bearing in mind that $\log$ has an inverse, namely, $\exp$. $$\log x^{1-\log x} = \log 1 \implies \cancel{\color{red}{\exp}}\left(\cancel{\color{red}{\log}} x^{1-\log x}\right) = \cancel{\color{blue}{\exp}}(\cancel{\color{blue}{\log}} 1) \implies x^{1-\log x} = 1.$$