Question

Логарифмическое ур-е: log 3 x+log 4 x=2 log 4 12

Answer 1

$log_3 x+log_4 x=2log_4 12;\ x>0;\ frac{log_4 x}{log_4 3}+log_4 x=2 log_4 (4*3);\ log_4 x (frac{1}{log_4 3}+1)=2*(log_4 4+log_4 3);\ log_4 x*frac{1+log_4 3}{log_4 3}=2*(1+log_4 3);\ log_4 x*frac{1}{log_4 3}=2;\ log_4 x=2*log_4 3;\ log_4 x=log_4 3^2;\ log_4 x=log_4 9;\ x=9$