Find the sum of all the values of $x$ satisfying ${log}_{10} (x + 9) + 2 {log}_{10} \sqrt{2 x - 1} = 1$ .

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Solution

${log}_{10} (x + 9) + 2 {log}_{10} \sqrt{2 x - 1} = 1$ is valid when $x + 9 > 0, 2 x - 1 > 0$
$\Rightarrow x > \frac{1}{2}$
Rewriting given equation as ${log}_{10} (x + 9) (2 x - 1) = 1, [∵ m log x = log x^{m} & log a + log b = log (a b)]$
$\Rightarrow (x + 9) (2 x - 1) = 10$
$\Rightarrow 2 x^{2} + 17 x - 19 = 0$
$\Rightarrow x = - 9.5, 1$
Since, $x > \frac{1}{2}$
Therefore, $x = 1$
and sum of roots is $1$ .
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