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If $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$, then
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Verified Answer
The correct answer is:
$-1 \leq x \leq 5$
Given, \(\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)\) if and only if
\(\begin{array}{l}
x^{2}-16 \leq 4 x-11 \\
\Rightarrow x^{2}-4 x-5 \leq 0 \\
x^{2}-5 x+x-5 \leq 0 \\
\Rightarrow x(x-5)+1(x-5) \leq 0 \\
(x-5)(x+1) \leq 0
\end{array}\)
Sign scheme of \(x^{2}-4 x-5 \leq 0\)
\(-1 \leq x \leq 5\)
\(\begin{array}{l}
x^{2}-16 \leq 4 x-11 \\
\Rightarrow x^{2}-4 x-5 \leq 0 \\
x^{2}-5 x+x-5 \leq 0 \\
\Rightarrow x(x-5)+1(x-5) \leq 0 \\
(x-5)(x+1) \leq 0
\end{array}\)
Sign scheme of \(x^{2}-4 x-5 \leq 0\)
\(-1 \leq x \leq 5\)
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