If $2 {log}_{e} 2 + {log}_{e} (2 x^{2} - 6 x + 5) = 2 {log}_{e} (2 x - 5)$ , then $x$ is equal to

only

\frac{1 + \sqrt{6}}{2}

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only

\frac{1 - \sqrt{6}}{2}

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Both A and B

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none of these

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Solution

The correct option is C none of these
Here, $2 x - 5 > 0$ and $2 x^{2} - 6 x + 5 > 0$
$\Rightarrow x > \frac{5}{2}$
$2 {log}_{e} 2 + {log}_{e} (2 x^{2} - 6 x + 5) = 2 {log}_{e} (2 x - 5)$
${log}_{e} 4 + {log}_{e} (2 x^{2} - 6 x + 5) = {log}_{e} (2 x - 5)^{2}$ ( $log x^{m} = m log x$ )
${log}_{e} (8 x^{2} - 24 x + 20) = {log}_{e} (4 x^{2} - 20 x + 25)$
$\Rightarrow 8 x^{2} - 24 x + 20 = 4 x^{2} - 20 x + 25$
$\Rightarrow 4 x^{2} - 4 x - 5 = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{6}}{2}$
Both of these values does not lie in the domain.
Hence, there is no value of $x$

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