[latex]log _{8}(x-2)-log _{8} (x-3) больше \frac{2}{3} [/latex]

log8_((x-2)/ (x-3)) > 2/3; 1/3 * log2_((x-2)/(x-3)) > 2/3;     *3 > 0; log2_((x-2)/(x-3)) > 2; log2_((x-2)(x-3)) > log2_4; 2> 1; ⇒ (x-2)/ (x-3) > 4; x-2/x-3  - 4  > 0; (x-2- 4x +12) / (x-3)  ) 0; (-3x+10)/(x-3) >0;  (3x-10) / (x-3) <0;      +            -            + _____(3)____(10/3)____x x∈(3; 10/3).  Сравним с одз. {x-2 >0; x-3 >0; ⇒ x > 3. Ответ х∈(3; 10/3)

[latex]log_8(x-2)-log_8(x-3)>\frac{2}{3},\; \; OOF:\; \left \{ {{x-2>0} \atop {x-3>0}} \right. \; \to \; x>3\\\\log_8\frac{x-2}{x-3}>log_88^{\frac{2}{3}},\; \; \; 8^{\frac{2}{3}}=\sqrt[3]{8^2}}=4\\\\\frac{x-2}{x-3}>4\\\\\frac{x-2-4x+12}{x-3}>0\\\\\frac{-3x+10}{x-3}>0\\\\\frac{3x-10}{x-3}<0,\; \; \; \; \; +++(3)---(\frac{10}{3})+++\\\\x\in (3,\frac{10}{3})[/latex]