Converting into alternate e-Form I was checking my results of an exercise online. The Maximum of my function was (1): $$\frac{4^{\frac{-1}{\log(2)}}}{\log^2(2)}$$
Now I wolframalpha tells me the alternate form is (2): $$\frac{1}{e^2\log^2(2)}$$
I feel like this form could be useful for future problems sometimes, but I absolutely have no clue how to get there. Could someone help me converting the maximum (1) into the alternate form (2)?
 A: Firstly we have $4=2^2$. So we get $$\frac{(2^2)^{\frac{-1}{\log(2)}}}{\log^2(2)}=\frac{2^{\frac{-2}{\log(2)}}}{\log^2(2)}$$
I´ve applied the rule $\left(\color{blue}a^{\color{red}b}\right)^{\color{green}c}=\color{blue}a^{\color{red}b\cdot \color{green}c}$. Furthermore it is  $2=e^{\log(2)}$.
$$=\frac{\left(e^{\log(2)}\right)^{\frac{-2}{\log(2)}}}{\log^2(2)}$$
Applying the rule above again and then cancelling $\log(2)$.
$$=\frac{e^{\log(2)\cdot {\frac{-2}{\log(2)}}}}{\log^2(2)}=\frac{e^{-2}}{\log^2(2)}=\frac{1}{e^2\cdot \log^2(2)}$$
A: It's a logarithmic rule-- a transformation rule, notice that $\log^2{(2)}$ appears common, so focus on $4^{\frac{-1}{\log{2}}}$
$$4^{\frac{-1}{\log{2}}} = 4^{-\frac{1}{\log{2}}}$$
$$ \frac{1}{\log_{a}{b}} = \log_{b}{a}$$
$$4^{-\frac{1}{\log{2}}} = 4^{-\log_{2}{e}}$$
$$4^{-\log_{2}{e}} = 2^{-2\log_{2}{e}}$$
$$2^{(\log_{2}{e})(-2)}$$
There's also another transformation rule
$$ a^{log_{a}{b}} = b$$
$$2^{(\log_{2}{e})(-2)} = (2^{\log_{2}{e}})^{-2}$$
$$e^{-2}$$
A: Let $x=4^{-1/\log2}$. Then
$$
\log x=-\frac{1}{\log2}\log4=-\frac{1}{\log2}\cdot 2\log2=-2
$$
which means $x=e^{-2}$.
A: Assuming  $\frac{4^{\frac{-1}{\ln2}}}{\ln^2(2)}$ = $\frac{1}{e^2\ln^22}$
$\frac{4^{\frac{-1}{\ln2}}}{1}$ = $\frac{1}{e^2}$
${4^{\frac{1}{\ln2}}}$ = ${e^2}$
$\frac{\ln4}{\ln2} $=2$\ln e$
$e$= $\frac{\ln 4}{2\ln2}$  =1
Reverse the above steps, we can prove  from (1) to (2).
