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 above expression to be defined,we must have,

$\{\begin{array}{c}{x}^2-16>0\\ 4x-11>0\end{array}$

$\Rightarrow \{\begin{array}{c}(x-4)(x+4)>0\\ x>\frac{11}{4}\end{array}$

$\Rightarrow \{\begin{array}{c}x<-4\text{or}x>4\\ x>\frac{11}{4}\end{array}$

So, taking common of both sets,

$\Rightarrow x\in (4,\infty )$ ...$\left(1\right)$

Now

${\log }_{10}(x^2-16)\le {\log }_{10}(4x-11)\Rightarrow x^2-16\le 4x-11\Rightarrow x^2-4x-5\le 0\Rightarrow x^2-5x+x-5\le 0\to (x-5)(x+1)\le 0$

$\Rightarrow x\in [-1,5]$ ...$\left(2\right)$
From (1) & (2) :

$\Rightarrow x\in (4,5]$

Ans: B