Выразите log300(2700) через m и n,если m=log2(3), n=log5(2) 1)m+n/mn 2)2n/3m...

$m=log_23$

$n=log_52$

$log_{300}2700= \frac{log_22700}{log_2300}=\frac{log_2(2^2\cdot3^3\cdot5^2)}{log_2(2^2\cdot3\cdot5^2)}=$

$\frac{log_22^2+log_23^3+log_25^2}{log_22^2+log_23+log_25^2}=\frac{2log_22+3log_23+2log_25}{2log_22+log_23+2log_25}=$

$\frac{2+3m+2 \cdot \frac{1}{log_52} }{2+m+2 \cdot \frac{1}{log_52} }=\frac{2+3m+\frac{2}{n} }{2+m+\frac{2}{n} }=\frac{\frac{2n+3mn+2}{n} }{\frac{2n+mn+2}{n} }=$

$\frac{2n+3mn+2}{n}:\frac{2n+mn+2}{n}=\frac{2n+3mn+2}{n} \cdot \frac{n}{2n+mn+2} = \frac{3mn+2n+2}{nm+2n+2}$

Ответ: 3) $\frac{3mn+2n+2}{nm+2n+2}$