Resposta:

[tex]\Large \textsf{Leia abaixo}[/tex]

Explicação passo a passo:

[tex]\Large \text{$ \sf log_x\:a = 8 $}[/tex]

[tex]\Large \text{$ \sf log_x\:b = 2 $}[/tex]

[tex]\Large \text{$ \sf log_x\:c = 1 $}[/tex]

[tex]\Large \text{$ \sf log_x\:\dfrac{a^3}{b^2\:.\:c^4} = 3\:.\:log_x\:a - 2\:.\:log_x\:b - 4\:.\:log_x\:c $}[/tex]

[tex]\Large \text{$ \sf log_x\:\dfrac{a^3}{b^2\:.\:c^4} = (3\:.\:8) - (2\:.\:2) - (4\:.\:1) $}[/tex]

[tex]\Large \text{$ \sf log_x\:\dfrac{a^3}{b^2\:.\:c^4} = 24- 4 - 4 $}[/tex]

[tex]\Large \boxed{\boxed{\text{$ \sf log_x\:\dfrac{a^3}{b^2\:.\:c^4} = 16 $}}}[/tex]

[tex]\Large \text{$ \sf log_x\:\dfrac{\sqrt[\sf 3]{\sf ab}}{c} = log_x\:\dfrac{(ab)^{\frac{1}{3}}}{c} $}[/tex]

[tex]\Large \text{$ \sf log_x\:\dfrac{\sqrt[\sf 3]{\sf ab}}{c} = \dfrac{1}{3}\:.\:(log_x\:a + log_x\:b)- log_x\:c $}[/tex]

[tex]\Large \text{$ \sf log_x\:\dfrac{\sqrt[\sf 3]{\sf ab}}{c} = \dfrac{1}{3}\:.\:(8 + 2)- 1 $}[/tex]

[tex]\Large \text{$ \sf log_x\:\dfrac{\sqrt[\sf 3]{\sf ab}}{c} = \dfrac{10}{3} - 1 $}[/tex]

[tex]\Large \boxed{\boxed{\text{$ \sf log_x\:\dfrac{\sqrt[\sf 3]{\sf ab}}{c} = \dfrac{7}{3} $}}}[/tex]