Question

log de 2 raiz de 2 na base 1/4

Answer 1

Ae meu brother,

use a definição de logaritmos:

$\Large\boxed{\log_a(c)=b~~\Rightarrow~~a^b=c}$

Vamos lá:

$\log_{ \tfrac{1}{4}}(2 \sqrt{2} )=x\\\\ \left( \dfrac{1}{4}\right)^x=2 \sqrt{2}\\\\ \left( \dfrac{1}{2^2} \right)^x= \sqrt[2]{2^2\cdot2}\\\\ (2^{-2})^x= \sqrt[2]{2^3}\\ \not2^{-2x}= \not2^{ \tfrac{3}{2} }\\\\ -2x= \dfrac{3}{2}\\\\ \Large\boxed{x=- \dfrac{3}{4} }$

$\log_2(0,25)=x\\\\ 2^x=0,25\\\\ 2^x= \dfrac{25}{100}\\\\ 2^x= \dfrac{25\div25}{100\div25}\\\\ 2^x= \dfrac{1}{4}\\\\ 2^x= \dfrac{1}{2^2}\\\\ \not2^x=\not2^{-2}\\\\ \huge\boxed{x=-2}$

TENHA ÓTIMOS ESTUDOS VLWW

Answer 2

Resolvendo a primeira:
$log _{ \frac{1}{4}} 2 \sqrt{2} = x \\ \\ ( 1/4)^{x} = 2 \sqrt{2} \\ (1/ 2^{2})^{x} = \sqrt{2* 2^{2} } \\ (2^{-2})^{x} = \sqrt{ 2^{3} } \\ (2)^{-2x} = 2^{ \frac{3}{2} }$
Igualando os expoentes:
$-2x = 3/2 \\ x=(3/2)/-2 \\ x= -3/4$

Resolvendo a segunda:
$log _{ 2}0,25 = x \\ \\ 2^{x} = 0,25 \\ 2^{x} = 1/4 \\ 2^{x} = (1/ 2^{2}) \\ 2^{x} = 2^{-2}$
Igualando os expoentes:
$x = -2$