Question

The solution set of the equation ${\log }_{1/5}(2x+5)+{\log }_5(16-x^2)\le 1$ is

• A. $(-\frac{5}{2},1)$
• B. $[-1,4)$
• C. $[1,-4]$
• D. $[-\frac{5}{2},4]$

Answer 1

The correct option is B $[-1,4)$

${\log }_{1/5}(2x+5)+{\log }_5(16-x^2)\le 1$

$\Rightarrow {\log }_5(16-x^2)-{\log }_5(2x+5)\le {\log }_{5}5[\because {\log }_{1/a}b=-{\log }_{a}b\text{\&}{\log }_{a}a=1]$
$\therefore \frac{16-x^2}{2x+5}\le 5or(16-x^2)\le 10x+25[\because \log a-\log b=\log \frac{a}{b}]$
or $x^2+10x+9\ge 0$
or $(x+9)(x+1)\ge 0$
$\therefore x\le -9$ or $x\ge -1$ ...(1)
Now by definition of $\log ,$ we must have
$2x+5>0$, i.e., $x>-\frac{5}{2}$
and $16-x^2>0$ or $x^2-16<0$
or $(x+4)(x-4)<0$
$\therefore -4<x<4$ and $x>-\frac{5}{2}$ ...(2)
Hence from (1) and (2), $x\ge -1$ and $x<4.$
$\therefore x\in [-1,4).$
Ans: B