If $x$ satisfies ${log}_{x} (2 x^{2} - 3 x - 4) = 2$ , find $x^{2}$

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Solution

${log}_{x} (2 x^{2} - 3 x - 4) = 2$

clearly above equation is valid when $2 x^{2} - 3 x - 4 > 0$
and $x > 0, \neq 1$

$x \in (\frac{3 + \sqrt{41}}{4}, \infty); \to (1)$

Rewriting above equation as $2 x^{2} - 3 x - 4 = x^{2}$

$\Rightarrow x^{2} - 3 x - 4 = 0$

$\Rightarrow x = 4, - 1$

Using $(1)$ , $x = 4$

Therefore, $x^{2} = 16$
Ans: 16

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