Question

Вычислить, срочно !!!

Answer 1

$\frac{log_224-\frac{1}{2}log_272}{log_318-\frac{1}{3}log_372}=\\\\\\=(log_224-log_272^{1/2}):(log_318-log_372^{1/3})=\\\\=log_2\frac{24}{\sqrt{72}}:log_3\frac{18}{\sqrt[3]{72}}=\\\\=log_2\frac{24}{6\sqrt{2}}:log_3\frac{18}{2\sqrt[3]{9}}=\\\\=log_2\frac{4}{\sqrt{2}}:log_3\frac{9}{\sqrt[3]{9}}=\\\\=log_2\frac{2^2}{2^{1/2}}:log_3\frac{3^2}{3^{1/3}}=\\\\=log_22^{2-1/2}:log_33^{2-1/3}=\\\\=(2-\frac{1}{2})*log_22:(2-\frac{1}{3})*log_33=\\\\=\frac{3}{2}:\frac{5}{3}=\frac{3}{2}*\frac{3}{5}=0,9$