Question
Solve for $x : \log( x - 1 ) + \log (x + 1 ) = \log_21.$
✓
Answer
$ \log (x-1)+\log (x+1)=\log _2 1$
$\Rightarrow \log (x-1)+\log (x+1)=0$
$\Rightarrow \log [(x-1)(x+1)]=0$
$\Rightarrow(x-1)(x+1)=1 \ldots .($ Since $\log 1=0)$
$\Rightarrow x^2-1=1$
$\Rightarrow x^2=2$
$\Rightarrow x= \pm \sqrt{2}$
$-\sqrt{2}$ can not be possible, since $\log$ of a negative number is not defined.
So,$x=\sqrt{2} \text {. }$
$\Rightarrow \log (x-1)+\log (x+1)=0$
$\Rightarrow \log [(x-1)(x+1)]=0$
$\Rightarrow(x-1)(x+1)=1 \ldots .($ Since $\log 1=0)$
$\Rightarrow x^2-1=1$
$\Rightarrow x^2=2$
$\Rightarrow x= \pm \sqrt{2}$
$-\sqrt{2}$ can not be possible, since $\log$ of a negative number is not defined.
So,$x=\sqrt{2} \text {. }$
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