Question

2log8(x-2) больше 2/3 + log8(x-3)

Answer 1

$2\log_{8}(x-2) > \dfrac{2}{3} + \log_{8}(x-3)\\\left \{ {\bigg{x - 2 > 0} \atop \bigg{x - 3 > 0}} \right. \ \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{x > 2} \atop \bigg{x > 3}} \right. \ \ \ \ \ \ \ \ \ \ x > 3$

$\log_{8}(x-2)^{2} > \log_{8}8^{2/3} + \log_{8}(x-3)\\\log_{8}(x-2)^{2} > \log_{8}4 + \log_{8}(x-3)\\\log_{8}(x-2)^{2} > \log_{8}4(x-3)\\(x-2)^{2} > 4(x-3)\\x^{2} - 4x + 4 > 4x - 12\\x^{2} - 8x + 16 > 0\\(x - 4)^{2} > 0;\\x \neq 4$

Ответ: $x \in (3; 4) \cup (4; +\infty)$