Question

Solve for x : log(x - 1) + log(x + 1) = log1
​

Answer 1

Answer:

+root 2 and -root 2

Step-by-step explanation:

Answer 2

$log(x-1)+log(x+1)=log1</p><p>\\</p><p>we\:know\:that\:log1=0</p><p>\\</p><p>So,</p><p>\\</p><p>log(x-1)+log(x+1)=0</p><p>\\</p><p>log(x-1)=\:-\:log(x+1)</p><p>\\</p><p>we\:know\:that\:x.log n=log n^x</p><p>\\</p><p>log(x-1)=log(x+1)^{-1}</p><p>\\</p><p>log(x-1)=log\left(\frac{1}{x+1}\right )</p><p>\\</p><p>x-1=\frac{1}{x+1}</p><p>\\</p><p>(x-1).(x+1)=1</p><p>\\</p><p>x^2-1=1</p><p>\\</p><p>x^2=2</p><p>\\</p><p>x=\pm\sqrt2</p><p>\\</p><p>Neglect - \sqrt2\:</p><p>\\</p><p>because\:\sqrt2\:is\:greater\:than\:1\:</p><p>\\</p><p>this\:leads\:negative\:values\:inside\:log\:</p><p>\\</p><p>this(never \:happens \:for\:real\:values</p><p>\\</p><p>So,\:the\:answer \:is\:x=\sqrt2.$

Or otherwise you can simply apply

log(ab)=log a + log b.

It results

log(x-1)+log(x+1)=log1

log[(x-1).(x+1)]=log1

(x-1)(x+1)=1

Again same approach

Hope it helps :))))