Converting into alternate e-Form

Question

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)?

Answer 1

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).

Answer 2

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 $\log(2)$.

$$=\frac{e^{\log(2)\cdot {\frac{-2}{\log(2)}}}}{\log^2(2)}=\frac{e^{-2}}{\log^2(2)}=\frac{1}{e^2\cdot \log^2(2)}$$

Answer 3

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 $$ a^{log_{a}{b}} = b$$ $$2^{(\log_{2}{e})(-2)} = (2^{\log_{2}{e}})^{-2}$$ $$e^{-2}$$

Answer 4

Let $x=4^{-1/\log2}$. Then $$ \log x=-\frac{1}{\log2}\log4=-\frac{1}{\log2}\cdot 2\log2=-2 $$ which means $x=e^{-2}$.