Question

log de 2 raiz de 2 na base 1/4

Answer 1

Usando da definição de ㏒ 

$\large\fbox{ y=\ell og_a(b)~\Longrightarrow~a^y=b }~~~~(0\ \textless \ a \neq 1)$

Portanto

$\large\begin{array}{l}y=\ell og_{\frac{1}{4}}(2\sqrt{2})~\Leftrightarrow~(\dfrac{1}{4})^y=2\sqrt{2}\end{array}$

Agora temos que ter em mente as seguintes propriedades exponenciais:

$\large\fbox{ a^{-1}=\dfrac{1}{a} }\\\\\large\fbox{ \sqrt[n]{a^m}=a^{\frac{m}{n}} }\\\\\large\fbox{ (a^m)^n=a^{m\cdot n} }$

Então

$\large\begin{array}{l}(\dfrac{1}{4})^y=2\sqrt{2}\\\\(4^{-1})^y=\sqrt{2^2\cdot2}\\\\4^{-y}=\sqrt{2^3}\\\\(2^2)^{-y}=2^{\frac{3}{2}}\\\\2^{-2y}=2^{\frac{3}{2}}\end{array}$

Agora nós temos uma igualdade entre potências de mesma base, o que implica em

$\large\fbox{ a^b=a^c~\Longleftrightarrow~b=c }~~(0\ \textless \ a \neq 1)$

Sendo assim

$\large\begin{array}{l}2^{-2y}=2^{\frac{3}{2}}~\Longleftrightarrow~-2y=\dfrac{3}{2}\\\\-2y=\dfrac{3}{2}\\\\y=\dfrac{3}{2}\cdot(-\dfrac{1}{2})\\\\\fbox{ y=-\dfrac{3}{4} }\\\\\fbox{ \ell og_{\frac{1}{4}}(2\sqrt{2})=-\dfrac{3}{4} }~~\leftarrow~~\text{resposta}\end{array}$