Question

calcule 2x log9 1 + [log0,5 (1/8)]^2​

Answer 1

Boa tarde!

$log_{9}1 = 0$

$log_{0.5}(\frac{1}{8})=log_{\frac{1}{2}}(\frac{1}{8})=\frac{log(\frac{1}{8})}{log(\frac{1}{2})}=\frac{log1- log8}{log1-log2}=\frac{0-log8}{0- log2}=\frac{-log8}{-log2}=\frac{- log{2}^{3}}{-log2}=\frac{3.(-log 2)}{-log2}=3$

2.0 + 3² = 0 + 9 = 9

Espero ter ajudado!