If ${log}_{(3 x - 1)} (x - 2) = {log}_{(9 x^{2} - 6 x + 1)} (2 x^{2} - 10 x - 2)$ , then $x$ equals.

9 - \sqrt{15}

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3 + \sqrt{15}

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2 + \sqrt{15}

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6 - \sqrt{5}

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Solution

The correct option is D $3 + \sqrt{15}$
${log}_{(3 x - 1)} (x - 2) = {log}_{(9 x^{2} - 6 x + 1)} (2 x^{2} - 10 x - 2)$
${log}_{(3 x - 1)} (x - 2) = {log}_{(3 x - 1)} \sqrt{2 x^{2} - 10 x - 2}$
$x - 2 = \sqrt{2 x^{2} - 10 x - 2}; x > 2$
Square
$x^{2} - 4 - 4 x = 2 x^{2} - 10 x - 2$
$x^{2} - 6 x - 6 = 0$
$x = \frac{6 \pm \sqrt{36 - 4 (- 6)}}{2}$
$x = \frac{6 \pm 2 \sqrt{15}}{2}$
$x = 3 - \sqrt{15}$ possible.

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