If ${log}_{10} (x + 1) + {log}_{10} (x - 1) = {log}_{10} 11 + 2 {log}_{10} 3$ , then value of $x$ will be

9

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10

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Solution

The correct option is D $10$
${log}_{10} (x + 1) + {log}_{10} (x - 1) = {log}_{10} 11 + 2 {log}_{10} 3$
${log}_{10} (x + 1) (x - 1) = log 11 \times 3^{2}$
$(x + 1) (x - 1) = 11 \times 3^{2}$
$x^{2} - 1 = 99$
$x^{2} = 100$
$x = \pm 10$
$∵ x$ cannot be negative.

Hence, $x = 10$

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Similar questions

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

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

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

{log}_{10} (x - 10) = 1

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If $l o g_{10} 5 + l o g_{10} (5 x + 1) = l o g_{10} (x + 5) + 1, t h e n (x^{2} - 9)$ is equal to:

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