Помогите пожалуйста разобраться с примерами

1. [latex]log_{x^2-5x+6}(x+3)=0\\log_{x^2-5x+6}(x+3)=log_{x^2-5x+6}(1)\\Potenciiryem\\x+3=1\\x=1-3\\x=-2[/latex] 2. [latex]log_{3-2x}(4x+3)=-1\\log_{3-2x}(4x+3)=-log_{3-2x}(3-2x)\\log_{3-2x}(4x+3)+log_{3-2x}(3-2x)\\log_{3-2x}((4x+3)(3-2x))=0\\log_{3-2x}((4x+3)(3-2x))=log_{3-2x}(1) \\Potenciiryem\\(4x+3)(3-2x)=1\\12x-8x^2+9-6x=1\\-8x^2+6x+9-1=0\\-8x^2+6x+8=0|:(-2)\\4x^2-3x-4=0\\D=(-3)^2-4*4*(-4)=9+64=73\\\\x_{1}=\frac{3+\sqrt{73}}{8}=\frac{3}{8}+\frac{\sqrt{73}}{8}\\\\x_{2}=\frac{3-\sqrt{73}}{8}=\frac{3}{8}-\frac{\sqrt{73}}{8}[/latex] 3. [latex]log_{4}(x)=2log_{4}(3)-3log_{4}(5)+log_{4}(35)\\log_{4}(x)=log_{4}(3^2)-log_{4}(5^3)+log_{4}(35)\\log_{4}(x)=log_{4}(9)-log_{4}(125)+log_{4}(35)\\log_{4}(x)=log_{4}(\frac{9}{125})+log_{4}(35))\\\\log_{4}(x)=log_{4}(\frac{9}{125}*35)\\\\Potenciityem\\\\x=\frac{9}{125}*35\\\\x=\frac{9*35}{125}\\\\x=\frac{9*7}{25}\\\\x=\frac{63}{25}\\\\x=2\frac{13}{25}[/latex]