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

9 - \sqrt{15}

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3 + \sqrt{15}

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2 + \sqrt{5}

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6 - \sqrt{5}

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Solution

The correct option is B $3 + \sqrt{15}$
${log}_{(3 x - 1)} (x - 2) = {log}_{(3 x - 1)^{2}} (2 x^{2} - 10 x - 2) {log}_{(3 x - 1)} (x - 2) = {log}_{(3 x - 1)} \sqrt{(2 x^{2} - 10 x - 2)}$
$(x - 2) = \sqrt{(2 x^{2} - 10 x - 2)} ∴ x > 2$
Squaring both the side
$x^{2} - 4 x + 4 = 2 x^{2} - 10 x - 2$
$x^{2} - 6 x - 6 = 0$
$x = \frac{6 \pm 2 \sqrt{15}}{2}$
$∴ x = 3 + \sqrt{15}$

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