Question

The solution set of the inequality ${\log }_{10}(x^2-16)\le {\log }_{10}(4x-11)$ is

• A. $(4,\infty )$
• B. $(4,5]$
• C. $(\frac{11}{4},\infty )$
• D. $(\frac{11}{4},5)$

Answer 1

The correct option is B $(4,5]$

${\log }_{10}(x^2-16)\le {\log }_{10}(4x-11)$
For the inequality to be defined,
$x^2-16>0\text{and}4x-11>0$
$\Rightarrow x\in (-\infty ,-4)\cup (4,\infty )\text{and}x>\frac{11}{4}$
So, $x\in (4,\infty )\cdots \left(1\right)$

Now, ${\log }_{10}(x^2-16)\le {\log }_{10}(4x-11)$
As the log function has base greater than $1$, so the inequality sign will remain same.
$x^2-16\le 4x-11\Rightarrow x^2-4x-5\le 0\Rightarrow (x-5)(x+1)\le 0\Rightarrow x\in [-1,5]\cdots \left(2\right)$

Hence, from $\left(1\right)$ and $\left(2\right)$,
$x\in (4,5]$