Question

7*log9(x^2-3x+2)<=8 + log9(x-2)^7/(x-1)

Answer 1

$7\cdot log_9(x^2-3x+2)\leq 8+log_9\dfrac{(x-2)^7}{x-1}\ \ ,\\\\\\ODZ:\ \left\{\begin{array}{l}x^2-3x+20\\\dfrac{(x-2)^7}{x-1}0\end{array}\right\ \ \left\{\begin{array}{l}(x-1)(x-1)0\\\dfrac{(x-2)^7}{x-1}0\end{array}\right\ \ \Rigyhtarrow \ \ x\in (-\infty ;\, 1\, )\cup (2;+\infty )\\\\\\log_9(x-2)^2(x-1)^7-log_99^8-log_9\dfrac{(x-2)^7}{x-1}\leq 0\ \ ,\ \ \ 0=log_91\ ,\\\\\\\dfrac{(x-2)^7(x-1)^7(x-1)}{9^8\cdot (x-2)^7}\leq 1\ \ ,\ \ \dfrac{(x-1)^8}{9^8}-1\leq 0\ \ ,\ \dfrac{(x-1)^8-9^8}{9^8}\leq 0\ ,$

$\star \ \ \ a^8-b^8=(a^4-b^4)(a^4+b^4)=(a-b)(a+b)(\underbrace {a^2+b^2}_{\geq 0})(\underbrace {a^4+b^4}_{\geq 0})\ \ \star \\\\\\(x-1-9)(x-1+9)\leq 0\\\\(x-10)(x+8)\leq 0\ \ \ ,\ \ \ \ +++[-8\, ]---[\, 10\, ]+++\\\\x\in [-8\, ;\, 10\, ]\\\\\\\left\{\begin{array}{l}x\in (-\infty ;1)\cup (2;+\infty )\\x\in [-8\, ;\, 10\, ]\end{array}\right\qquad \Rightarrow \quad x\in [-8\, ;1\, )\cup (\, 2\, ;10\, ]$