Question

What is the solution set of the inequality log base e (x2 – 16) ≤ log base e (4x – 11)?​

Answer 1

Step-by-step explanation:

We have,

$\log_{e}( {x}^{2} - 16 ) \leqslant \log_{e}(4x - 11)$

$\implies {x}^{2} - 16 \leqslant 4x - 11$

$\implies {x}^{2} - 4x - 16 - 11 \leqslant 0$

$\implies {x}^{2} - 4x - 27 \leqslant 0$

$\implies \{x - (2 + \sqrt{31} ) \} \{ x - (2 - \sqrt{31} )\} \leqslant 0$

$\implies x \in [ 2 - \sqrt{31} \: , \: 2 + \sqrt{31} ]$

Answer 2

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