Question

If ${\log }_{1/2}\left({\log }_8\frac{x^2-2x}{x-3}\right)<0$ then $x\in \left(a_1,a_2\right)\cup \left(a_3,a_4\right)$. The value of $\left(a_1+a_2+a_3\right)$ equals

• A. 10
• B. 11
• C. 12
• D. 13

Answer 1

Answer: (D)

${\log }_{1/3}\left({\log }_8\frac{x^2-2x}{x-3}\right)<0;{\log }_8\left(\frac{x^2-2x}{x-3}\right)>1;\frac{x^2-2x}{x-3}>8\Rightarrow \frac{(x-6)(x-4)}{x-3}>0$
$\Rightarrow x\in (3,4)\cup (6,\infty )\Rightarrow a_1+a_2+a_3=13$