Question

sabem fazer essa? de Logaritmo

Answer 1

Faz assim Elisandra ...

$\frac{Log_{3}1+Log_{10}0,01}{Log_{2\frac{1}{64}}*Log_{4\sqrt{8}}}$

$\frac{Log_{3}1+Log_{10}0,01}{Log_{2\frac{1}{64}}*Log_{4\sqrt{8}}}\\\\ \bullet {} Log_{3}1 = 3^{x} = 1 \to 0;\\\\ \bullet {}Log_{10}0,01 = Log_{10}10^{-2}= -2(Log_{10}10) = 10^{x} = 10 \to -2.1=-2;\\\\ \bullet {}Log_{2} \frac{1}{64}=Log_{2}2^{-6}= -6(Log_{2}2) = 2^{x}=2 \to -6.1=-6;\\\\ \bullet {}Log_{4} \sqrt{8}= Log_{4}2^{\frac{3}{2}} =\frac{3}{2}(Log_{4}2)=4^{x}=2= \to \frac{3}{2}*\frac{1}{2}=\frac{3}{4};\\\\ \text{Agora, basta substituir os valores ...}\\\\\frac{0 - 2}{-6*\frac{3}{4}}=\boxed{\frac{4}{9}}$

Até mais !