Question

Log8 log32 1024-log4 3(над корнем) √2

Answer 1

$log_8\, log_{32}\, 1024-log_4\sqrt[3]{2}=log_8\, log_{2^5}2^{10}-log_{2^2}\, 2^{1/3}=\\\\=log_8\, (\frac{1}{5}\cdot 10\cdot \underbrace {log_22}_{1})-\frac{1}{2}\cdot \frac{1}{3}\underbrace {log_22}_{1}=log_{2^3}\, 2-\frac{1}{6}=\\\\=\frac{1}{3}- \frac{1}{6}=\frac{2-1}{6}=\frac{1}{6}$