questão 1

[tex]\displaystyle \sf \log_{1/2}(x-1)+\log_{1/2}(3x-2)\geq -2 \\\\ \text{Condi\c c\~ao de exist\^encia (C.E ) :}\\\\ I)\ x-1 > 0 \to x > 1 \ ; \\\\ II)\ 3x-2 > 0 \to x > \frac{2}{3}\ ; \\\\\ CE = (I)\cap (II) \\\\ CE=(x > 1)\cap \left(x > \frac{2}{3}\right)\\\\\\ \boxed{\sf C.E = x > 1}[/tex]

temos :

[tex]\displaystyle \sf \log_{1/2}(x-1)+\log_{1/2}(3x-2)\geq -2 \\\\ \log_{(2^{-1})}\left[(x-1)(3x-2)\right]\geq -2\\\\ -1\cdot \log_{2}(3x^2-5x+2) \geq -2 \ \ \cdot (-1) \\\\ \log_2(3x^2-5x+2) \leq 2\\\\ C.E_2 :\\\\ 3x^2-5x+2 > 0 \\\\ \text{ra\'izes} \\\\ x=\frac{-(-5)\pm\sqrt{(-5)^2-4\cdot 3\cdot 2}}{2\cdot 3}\\\\\\ x= \frac{5\pm1}{6} \to x = 1\ e \ x =\frac{2}{3}\\\\\ \text{par\'abola c/ concavidade p/cima, negativa entre as ra\'izes. Ent\~ao :}\\\\ \boxed{\sf CE_2 : x < \frac{2}{3}\ e \ x > 1 }\\\\\ temos :[/tex]

[tex]\displaystyle \sf \log_2(3x^2-5x+2)\leq 2\\\\ 3x^2-5x+2\leq 2^2 \\\\ 3x^2-5x+2\leq 4\\\\ 3x^2-5x-2\leq 0 \\\\ ra\'izes }: \\\\ x=\frac{-(-5)\pm\sqrt{(-5)^2-4\cdot 3\cdot (-2)}}{2\cdot 3}\\\\\ x=\frac{5\pm\sqrt{25+24}}{6}=\frac{5\pm\sqrt{49}}{6}\\\\\\ x=\frac{5\pm 7}{6} \\\\ x=\frac{5+7}{6} \to x=\frac{12}{6} \to x= 2 \\\\\\ x=\frac{5-7}{6} \to x=\frac{-2}{6} \to x=\frac{-1}{3}[/tex]

Parábola com concavidade para cima e negativa entre as raízes. Então :

[tex]\displaystyle \sf \left(\frac{-1}{3}\leq x\leq 2\right)\\\\\ \text{Solu\c c\~ao }:\\\\ S = (C.E_1)\cap(C.E_2)\cap \left(\frac{-1}{3}\leq x\leq 2\right)\\\\\\ S=(x > 1)\cap\left(x < \frac{2}{3}\right)\cap \left(\frac{-1}{3}\leq x\leq 2\right) \\\\\\\ \large\boxed{\sf \ S :\left\{ x\in\mathbb{R}\ |\ 1 < x\leq 2\ \right\}\ }\checkmark[/tex]

questão 2

[tex]\displaystyle \sf \log_{1/2}x+\log_{1/2}(x-2) > -3 \\\\ \log_{2^{-1}}+\log_{2^{-1}}(x-2) > -3 \\\\\ -\log_2x-\log_2(x-2) > -3\ \cdot (-1) \\\\ log_2x+\log(x-2) < 3 \\\\ C.E_1: \\\\ x > 0 \ ; \ x-2 > 0 \to x > 2\\\\ \boxed{\sf C.E_1:x > 2} \\\\\ temos :\\\\ \log_2x+\log_2(x-2) < 3 \\\\ \log_2(x(x-2)) < 3 \\\\ \log_2(x^2-2x) < 3 \\\\ C.E_2 : \\\\ x^2-2x > 0 \\\\ \text{ra\'izes :}\ x=0 \ e \ x = 2 \\\\ \text{par\'abola com concavidade p/cima, negativa entre as ra\'izes. Ent\~ao }: \\\\ C.E_2 : x < 0\ e \ x > 2[/tex]

[tex]\displaystyle \sf temos : \\\\ \log_2(x^2-2x) < 3\\\\ x^2-2x < 2^3 \\\\ x^2-2x < 8\\\\ x^2-2x+1 < 9 \\\\ (x-1)^2 < 3^2\\\\ |x-1| < 3 \\\\ x-1 < 3 \to x < 4 \\\\ -(x-1) < 3 \to x-1 > -3 \to x > -2 \\\\ -2 < x < 4 \\\\ \text{solu\c c\~ao}: \\\\ S=C.E_1\cap C.E_2\cap (-2 < x < 4) \\\\ S=(x > 2)\cap(x < 0)\cap (-2 < x < 4) \\\\ \boxed{\sf \ S =\left\{x\in\mathbb{R}\ |\ 2 < x < 4\ \right\}\ }\checkmark (d)[/tex]