Question

Opções:
a)1
b)5
c)3
d)2
e)4​

Answer 1

$\frac{1}{\log_x8}+\frac{1}{\log_{2x}8}+\frac{1}{\log_{4x}8}=2$

$1/\frac{\log8}{\log x}+1/\frac{\log8}{\log 2x}+1/\frac{\log8}{\log 4x}=2$

$\frac{\log x}{\log 8}+\frac{\log 2x}{\log 8}+\frac{\log 4x}{\log8}=2$

$\frac{\log x+\log2x+\log4x}{\log8}=2$

$\log x+\log2x+\log4x=2\log8$

$\log(x.2x.4x)=\log8^2$

$x.2x.4x=8^2$

$8x^3=64$

$x^3=\frac{64}{8}$

$x^3=8$

$x=\sqrt[3]{8}$

$x=2$

Apenas 1 condição de existência: x=2

Gabarito: a)