Вопрос в картинках...

Question

\frac{2}{3} " alt="log _{8}(x-2)-log _{8} (x-3)> \frac{2}{3} " align="absmiddle" class="latex-formula">

Answer 1

\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})" alt="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})" align="absmiddle" class="latex-formula">

Comments

спасибо

Answer 2

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;<br>     +            -            +
_____(3)____(10/3)____x
x∈(3; 10/3).

 Сравним с одз.
{x-2 >0;
x-3 >0; ⇒ x > 3.
Ответ х∈(3; 10/3)