Question

qual é o Log1/4 base 2 raiz de 2?

Answer 1

Pela definição y = log_{a}(b) ⇔ a^{y} = b

∴

$\large\begin{array}{l}y=\ell og_{\frac{1}{4}}(2\sqrt{2})~\Leftrightarrow~(\frac{1}{4})^y=2\sqrt{2}\\\\(4^{-1})^y=2\cdot2^{\frac{1}{2}}\\\\4^{-y}=2^{1+\frac{1}{2}}\\\\(2^2)^{-y}=2^{\frac{3}{2}}\\\\2^{-2y}=2^{\frac{3}{2}}\end{array}$

Já que temos uma igualdade entre potências de mesma base podemos igualar os expoentes.

$-2y=\dfrac{3}{2}\\\\\\y=\dfrac{~~3}{\dfrac{~~2}{-2}}\\\\\\y=-\dfrac{3}{4}\\\\\\\fbox{ \ell og_{\frac{1}{4}}(2\sqrt{2})=-\dfrac{3}{4} }~~\leftarrow~~\text{resposta}$

Answer 2

Explicação passo-a-passo:

olá

resolução

$log\frac{1}{4} \sqrt{2} = x = > ( \frac{1}{ {}^{4} }) {}^{x} = 2 \sqrt{2} \\(\frac{1}{4} {}^{} ) {}^{x} = 2 \sqrt{2} = > ( \frac{1}{4} ) {}^{x} = {2}^{1} .2 {}^{ \frac{1}{2} } = > \\ (2 { }^{ - 2 {}^{} } ) {}^{x} = 2 {}^{1 + \frac{1}{2} } = > 2 {}^{ - 2x} = 2 {}^{ \frac{3}{2} } = > - 2x \\ = \frac{3}{2} = > x = - \frac{3}{4}$

logo , log1/4 2 raiz de 2 = -3/4

bons resultados