Question

log x na base 4 + log x na base 8 - log x na base 2= -1

Answer 1

Explicação passo-a-passo:

Equação Logarítmica:

$\mathsf{\log_{4}x+\log_{8}x-\log_{2}x~=~-1 }$

$\mathsf{\dfrac{\log_{2}x}{\log_{2}4}+\dfrac{\log_{2}x}{\log_{2}8}-\dfrac{\log_{2}x}{\log_{2}2}~=~-1}$

$\mathsf{\dfrac{\log_{2}x}{2}+\dfrac{\log_{2}x}{3}~=~-1+\dfrac{\log_{2}x}{1} }$

$\mathsf{\dfrac{3\log_{2}x+2\log_{2}x}{6}~=~-1+\dfrac{\log_{2}x}{1} }$

$\mathsf{\dfrac{5\log_{2}x}{6}~=~-1+\dfrac{\log_{2}x}{1} }$

$\mathsf{\dfrac{5\log_{2}x}{6}-\dfrac{\log_{2}x}{1}~=~-1 }$

$\mathsf{\dfrac{5\log_{2}x-6\log_{2}x}{6}~=~-1 }$

$\mathsf{\dfrac{-\log_{2}x}{6}~=~-1 }$

$\mathsf{-\log_{2}x~=~-1.6 }$

$\mathsf{-\log_{2}x~=~-6 }$

$\mathsf{\log_{2}x~=~+6 }$

$\mathsf{2^6~=~x }$

$\boxed{\boxed{\mathsf{x~=~64 }}}}$$\checkmark$

Espero ter ajudado bastante!)