Question

The solution set of the inequality ${\log }_{\pi }(x+27)-{\log }_{\pi }(16-2x)<{\log }_{\pi }x$ is

• A. $x\in (3,\frac{9}{2})\cup (8,\infty )$
• B. $x\in (3,8)$
• C. $x\in (3,\frac{9}{2})\cup (\frac{9}{2},8)$
• D. $x\in (3,\frac{9}{2})$

Answer 1

The correct option is D $x\in (3,\frac{9}{2})$

Given,
${\log }_{\pi }(x+27)-{\log }_{\pi }(16-2x)<{\log }_{\pi }x$
$\log$ function is defined when
$\left(i\right)x+27>0$
$\Rightarrow x\in (-27,\infty )$
$\left(ii\right)16-2x>0$
$\Rightarrow x\in (-\infty ,8)$
$\left(iii\right)x>0$
$\Rightarrow x\in (0,\infty )$
From $\left(i\right),\left(ii\right)$ and $\left(iii\right)$, we get
$x\in (0,8)\cdots \left(iv\right)$

${\log }_{\pi }(x+27)-{\log }_{\pi }(16-2x)<{\log }_{\pi }x$
$\Rightarrow {\log }_{\pi }\frac{(x+27)}{(16-2x)}<{\log }_{\pi }x$
$\Rightarrow \frac{x+27}{16-2x}<x$
$\Rightarrow \frac{x+27}{16-2x}-x<0$
$\Rightarrow \frac{2x^2-15x+27}{16-2x}<0$
$\Rightarrow \frac{(2x-9)(x-3)}{x-8}>0$
By wavy curve method
$\Rightarrow x\in (3,\frac{9}{2})\cup (8,\infty )\cdots \left(v\right)$

From $\left(iv\right)$ and $\left(v\right),$ we get
$x\in (3,\frac{9}{2})$