Question
Solve the following:$\log (x + 1) + \log (x - 1) = \log 48$
✓
Answer
$\log (x + 1) + \log (x - 1) = \log 48$
$\Rightarrow \log {(x + 1)(x - 1)} = \log 48$
$\Rightarrow \log (x^2 - 1) = \log 48$
$\Rightarrow x^2- 1 = 48$
$\Rightarrow x^2 = 49$
$\Rightarrow x = 7 ...($neglecting the negative value$).$
$\Rightarrow \log {(x + 1)(x - 1)} = \log 48$
$\Rightarrow \log (x^2 - 1) = \log 48$
$\Rightarrow x^2- 1 = 48$
$\Rightarrow x^2 = 49$
$\Rightarrow x = 7 ...($neglecting the negative value$).$
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