Question

log raiz cúbica 64 na base 2 - log 1 na base 8 + log 27/64 na base 4/3

Answer 1

Log2 ³√64 - Log8 1 + Log4/3 27/64

Log2 ³√4³ - 0 + Log4/3 3³/4³

Log2 4 + Log4/3 (3/4)³

Log2 2^(2) + Log4/3 (4/3)^(-3)

(2) + (-3)

(2) - (3)

-1 ✓

Answer 2

Olá!

$log_2 \: \sqrt[3]{64} - log_8 \: 1 + log_{\frac{4}{3} } \: \frac{27}{64} \\ \\ = log_2 \: \sqrt[3]{2^6} - log_{ \cancel{8} } \: \cancel{8}^0 + log_{\frac{4}{3} } \: (\frac{3}{4})^3 \\ \\= log_{ \cancel{2} } \: \cancel{2}^{ \frac{6}{3} } - 0+ log_{ \cancel{\frac{4}{3} } } \: \cancel{(\frac{4}{3})}^{-3} \\ = 2 - 0 -3 \\ = 2-3 \\ = -1$

Boa interpretação;