Find the value of $x$ which satisfy the equation ${log}_{2} (x^{2} - 3) - {log}_{2} (6 x - 10) + 1 = 0$

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Solution

${log}_{2} (x^{2} - 3) - {log}_{2} (6 x - 10) + 1$ is valid when $x^{2} - 3 > 0, 6 x - 10 > 0$
$\Rightarrow x > \sqrt{3}$
Rewrite given equation as,
${log}_{2} (\frac{x^{2} - 3}{6 x - 10}) = - 1 \Rightarrow \frac{x^{2} - 3}{6 x - 10} = \frac{1}{2}$
$\Rightarrow x^{2} - 3 = 3 x - 5 \Rightarrow x^{2} - 3 x + 2 = 0 \Rightarrow x = 1, 2$
Since, $x > \sqrt{3}$
Thus, $x = 2$
Ans: 2

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