If log _{e}left(x^{2}-16right) leq log _{e}(4 | vedclass

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(A) For the logarithmic inequality $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$ to be defined, the arguments must be positive: $1) \ x^{2}-

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$4 < x \leq 5$

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(A) For the logarithmic inequality $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$ to be defined, the arguments must be positive: $1) \ x^{2}-16 > 0$ $\Rightarrow (x-4)(x+4) > 0$ $\Rightarrow x \in (-\infty, -4) \cup (4, \infty)$ $2) \ 4x-11 > 0 \Rightarrow x > \frac{11}{4} = 2.75$ Combining these, the domain is $x > 4$. Now, solving the inequality: $x^{2}-16 \leq 4x-11$ $x^{2}-4x-5 \leq 0$ $(x-5)(x+1) \leq 0$ This holds for $x \in [-1, 5]$. Taking the intersection of the domain $(x > 4)$ and the solution $(x \in [-1, 5])$, we get $4 < x \leq 5$.

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

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