Question

ещё одна задача с логарифмами во вложениями

Answer 1

$\boxed{\;\boldsymbol{\big8}\;}\\\\\\\dfrac{log_345}{log_53}-\dfrac{log_315}{log_{15}3}=\dfrac{log_39+log_35}{\dfrac{1}{log_35}}-\dfrac{log_3{15}}{\dfrac{1}{log_3{15}}}=log_35*\Big(2+log_35\Big)-log^2_3{15}=\\\\\\=2*log_35+log^2_35-\Big(log_33+log_35\Big)^2=2*log_35+log^2_35-\Big(1+log_35\Big)^2=\\\\\\=2*log_35+log^2_35-\Big(1+2*log_35+log^2_35\Big)=2*log_35+log^2_35-1-\\\\\\-2*log_35-log^2_35=-1$

$Formyla\;:\\\\\boxed{\;\;log_a\Big(bc\Big)=log_ab+log_ac\;\;}\\\\\\\boxed{\;\;log_ab=\dfrac{1}{log_ba}\;\;\;}\\\\\\Otvet\;:\;\;-1$