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

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Solution

The correct option is B

$l o g_{(3 x - 1)} (x - 2) = l o g_{(9 x^{2} - 6 x + 1)} (2 x^{2} - 10 x - 2) l o g_{(3 x - 1)} (x - 2) = l o g_{(3 x - 1)} \sqrt{2 x^{2} - 10 x - 2} x - 2 = \sqrt{2 x^{2} - 10 x - 2}$ x>2
Square
$x^{2} - 4 - 4 x = 2 x^{2} - 10 x - 2 x^{2} - 6 x - 6 = 0 x = \frac{6 \pm \sqrt{36 - 4 (- 6)}}{2} x = \frac{6 \pm 2 \sqrt{15}}{2} x = 3 + \sqrt{15} is the possible solution$

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