Question

If $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$, then

• A. $4
• B. $x\, 4$
• C. $-1 \le x\le5$
• D. $x\, 5$

Answer 1

Answer: (A)

$4

Answer 2

Given, $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$ if and only if $x^{2}-16 \leq 4 x-11$ $\Rightarrow x^{2}- 4 x-5 \leq 0$ $x^{2}-5 x+x-5 \leq 0$ $\Rightarrow x(x-5)+1(x-5) \leq 0$ $(x-5)(x+1) \leq 0$ Sign scheme of $x^{2}-4 x-5 \leq 0$ $-1 \leq x \leq 5$