Даю 20 баллов. помогите решить

Решите задачу:

$lg(3x-1)=lg(7-2x) \\ 3x-1=7-2x \\ 5x=8 \\ x =8/5 ; x=1,6 \\ \\ log_{0,5}(x^2-4x+20)=-5 \\ x^2-4x+20=2^5 \\ x^2-4x-12=0 \\ D=16-4*(-12)*1=64 \\ x_{1,2} = \frac{4+/-8}{2} \\ x_1 = 6 ; x_2 = -2 \\ x^2-4x+20\ \textgreater \ 0 (always \ \textgreater \ 0) \\ \\ log_3(x+2)+log_3(x+3)=log_3(-2x) \\ log_3(x+2)*(x+3)=log_3(-2x) \\ log_3(x^2+3x+2x+6)=log_3(-2x) \\ log_3(x^2+5x+6)=log_3(-2x) \\ x^2+5x+6+2x=0 \\ x^2+7x+6=0 \\ D=49-4*6*1=49-24=25 \\ x_{1,2} = \frac{-7+/-5}{2} \\ x_1 = -6 ; x_2 = -1 \\ -2x\ \textgreater \ 0 ; x\ \textless \ 0 \\ x+2\ \textgreater \ 0 x\ \textgreater \ -2 \\$
$x=-1 \\ lg^2(x)-lg(x)-2=0 \\ x\ \textgreater \ 0 ; \\ D=1-4*(-2)*1=9 \\ x_{1,2} = \frac{1+/-3}{2} = \\ x_1 = 2 ; x_2 = -1 \\ lgx=2 ; lg = 100 ; lgx=-1 x=0,1 \\ \\ log_8(27x-1)+log_{27x-1}(8)=10/3 \\ 27x-1*8=10/3 \\ 27x-8=10/3 \\ 27x=10/3+8 \\ 27x=34/3 \\ x=34/3/3 \\ x=34/81$