Question
Solve the following:$\log ( x + 1) + \log ( x - 1) = \log 11 + 2 \log 3$

Answer

$\log ( x + 1) + \log ( x - 1) = \log 11 + 2 \log 3$
$\Rightarrow \log [(x + 1)(x - 1)] = \log 11 + \log 3^2$
$\Rightarrow \log {x^2 - 1} = \log (11.9)$
$\Rightarrow \log {x^2 - 1} = \log99$
$\Rightarrow x^2 - 1 = 99$
$\Rightarrow x^2 = 100$
So, $x = 10$ or $-10$
Negative value is rejected
So, $x = 10.$

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