Question

1. log(2)56-log(2)7+16 log(2)3 2. log(7)(3x-5)-log(7)(9-2x)=1 3. 4-lg^2x=3tgx 4. log 1\3(2-3x)< -2 ренить систему log(2)(x-y)+2log(4)(x-y)=3 3(2)+log(3)(2x-y)=45

Answer 1

$log_256-log_27+16 log_23= log_2(56/7)+log_2(3^1^6)= \\ =log_28+log_2(3^1^6)=3+16log_23 \\ \\ 2.\\ log_7(3x-5)-log_7(9-2x)=1 \\log_7\frac{3x-5}{9-2x}=log_77 \\\ \frac{3x-5}{9-2x}=7 \\ \frac{3x-5-7(9-2x)}{9-2x}=0 \\ \frac{3x-5-63+14x}{9-2x}=0 \\ 17x=68 = x=4\\ \\ 3.\\ 4-lg^2x=3lgx \\ t=lgx\\ 4-t^2-3t = 0 \\ t^2+3t-4=0 \\ (t+4)(t-1)=0\\lgx_1=1 <= x=10 \\ lgx_2=-4 <= x=10^{-4} \\ \\ 4.\\ log _{1/3}(2-3x)<-2 \\ -log _{3}(2-3x)<-log_39 \\ log _{3}(2-3x)log_39 \\ (2-3x)9 \\ 3x<11 = x \in (- \infty; 11/3)$