Решите пожалуйста логарифмы 10 класс

Question 1

Question 2

Решите задачу:

$2log_327^{-1}+6^{log_672-log_62}=2log_33^{-3}+6^{log_636}=-6+36=30\\ 3lg5+lg8=lg5^3+lg8=lg(125*8)=lg1000=lg10^3=3\\ \\ log_{0.1}(x^2-3x)=-1\\log_{0.1}(x^2-3x)=log_{0.1}0,1^{-1}\\x^2-3x\ \textgreater \ 0\\ x(x-3)\ \textgreater \ 0\\ x\in (-\infty;0)U(3;+\infty)\\x^2-3x=10\\x^2-3x-10=0\\D=3^2+4*10=49\\ x_1= \frac{3+7}{2} =5, \ x_2= \frac{3-7}{2} =-2\\ Otvet: \ x=5, \ x=-2\\ \\ 2log_5(-x)=log_5(x+2)\\ -x\ \textgreater \ 0, \ x+2\ \textgreater \ 0\\x\ \textless \ 0, \ x\ \textgreater \ -2\\x\in(-2;0)\\ log_5(-x)^2=log_5(x+2)\\x^2=x+2\\ x^2-x-2=0\\ x=-1,\ x=2\\ Otvet: \ x=-1\\ \\$

$log_{0,2}(3x-1) \geq log_{0,2}(3-x)\\ 3x-1\ \textgreater \ 0, \ 3-x\ \textgreater \ 0\\ 3x\ \textgreater \ 1, \ x\ \textless \ 3\\x\ \textgreater \ \frac{1}{3}, \ x\ \textless \ 3\\ x\in( \frac{1}{3} ;3)\\ 0,2\ \textless \ 1 \ \to \ 3x-1 \leq 3-x\\ 3x+x \leq 3+1\\ 4x \leq 4\\ x \leq 1\\Otvet: \ x\in ( \frac{1}{3} ;1]\\ \\ log_3(x^2-1)\ \textless \ log_3(x+1)+1\\ x^2-1\ \textgreater \ 0, \ x+1\ \textgreater \ 0\\ x^2\ \textgreater \ 1, \ x\ \textgreater \ -1 \\ x\in(-\infty;-1)U(1;+\infty), \ x\ \textgreater \ -1\\ x\in(1;+\infty)\\ log_3(x^2-1)\ \textless \ log_3(x+1)+log_33\\ log_3(x^2-1)\ \textless \ log_3(3x+3)\\ 3\ \textgreater \ 1\ \to \ x^2-1\ \textless \ 3x+3\\ x^2-3x-4\ \textless \ 0\\ x=-1, \ x=4\\ (x+1)(x-4)\ \textless \ 0\\ x\in(-1;4)\\ Otvet: \ x\in(1;4)$