The value of $x$ if ${log}_{10} x^{2} - 7 x - 6 = 1 - {log}_{10} 5$

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Solution

The correct option is D $8$
Given:
${log}_{10} x^{2} - 7 x - 6 = 1 - {log}_{10} 5$
For the $log$ to be defined,
$x^{2} - 7 x - 6 > 0$
So, the roots of equation
$x^{2} - 7 x - 6 = 0 \Rightarrow x = \frac{7 \pm \sqrt{73}}{2}$
So,
$x \in (- \infty, \frac{7 - \sqrt{73}}{2}) \cup (\frac{7 + \sqrt{73}}{2}, \infty) \dots (1)$

Now,
${log}_{10} x^{2} - 7 x - 6 = 1 - {log}_{10} 5 \Rightarrow {log}_{10} x^{2} - 7 x - 6 + {log}_{10} 5 = 1 \Rightarrow {log}_{10} 5 (x^{2} - 7 x - 6) = 1 \Rightarrow 5 (x^{2} - 7 x - 6) = 10$
$\Rightarrow x^{2} - 7 x - 6 = 2$
$\Rightarrow x^{2} - 7 x - 8 = 0$
$\Rightarrow x^{2} - 8 x + x - 8 = 0$
$\Rightarrow (x + 1) (x - 8) = 0$
$\Rightarrow x = - 1, 8$
From equation $(1)$ , both values are accepted
$∴ x = - 1, 8$

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