Find $x$ , if ${log}_{10} (x + 1) + {log}_{10} (x - 1) = {log}_{10} 11 + 2 {log}_{10} 3$

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Solution

we know that
$l o g_{10} a^{n} = n l o g_{10} a$

$l o g_{10} a b = l o g_{10} a + l o g_{10} b$

So $l o g_{10} (x + 1) + l o g_{10} (x - 1) = l o g_{10} 11 + 2 l o g_{10} 3$

$l o g_{10} (x + 1) (x - 1) = l o g_{10} (11 * 3^{2})$

so $x^{2} - 1 = 99 ⟹ x^{2} = 100$

$x = 10$

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