Question

If $x$ satisfies the equation $\log (1+4x^2-4x)-\frac{1}{2}\log (19+x^2)=\log (1-2x)$, then find $\left|x\right|$

Answer 1

$\log (1+4x^2-4x)-\frac{1}{2}\log (19+x^2)=\log (1-2x)$
Clearly $x<\frac{1}{2}$.
Rewrite the equation as
$\log {(1-2x)}^2-\frac{1}{2}\log (19+x^2)=\log (1-2x)[\because 1+4x-4x^2=(1-2x{)}^2]$
$\Rightarrow 2\log (1-2x)=\log (19+x^2\left)\right[\because \log a^m=m\log a]$
$\Rightarrow 1-4x+4x^2=19+x^2\Rightarrow x=\frac{2-\sqrt{58}}{3}$

$⟹x=-1.87$
Ans: 2