Question

O valor da expressão log2 - log 1/0,0001+ log5 + $log ^{} _{ \sqrt{2} } 8$é igual a

Answer 1

Olá, Jade.

$\log2 - \log \frac1{0,0001}+ \log5 + \log_{ \sqrt{2} } 8= \\\\ =\log2 - \log \frac1{10^{-4}}+ \log5 + \log_{ \sqrt{2} } {(\sqrt2)^6}= \\\\ =\log2 - \log 10^{4}+ \log5 + 6= \\\\ =\log2 - 4+ \log5 + 6= \\\\ =\log(2\cdot5)+2= \\\\ =\log10+2= \\\\ =1+2= \\\\ =3$