Question

Find sum of all integer values of $x$ satisfying : ${\log }^2(4-x)+\log (4-x).\log (x+\frac{1}{2})-2{\log }^2(x+\frac{1}{2})=0$

Answer 1

${\left(\log (4-x)\right)}^2+\log (4-x).\log (x+\frac{1}{2})-2{\left(\log (x+\frac{1}{2})\right)}^2=0$

Is defined when $4-x\ge 0\Rightarrow x\le 4$ and $x+\frac{1}{2}\ge 0\Rightarrow x\le -\frac{1}{2}$

$\Rightarrow \left(\log (4-x)+2\log (x+\frac{1}{2})\right).\left(\log (4-x)-\log (x+\frac{1}{2})\right)=0$
$\Rightarrow \left(\log (4-x)+{(x+\frac{1}{2})}^2\right).\left(\log \left(\frac{4-x}{x+\frac{1}{2}}\right)\right)=0$
$\Rightarrow 4-x+(x^2+\frac{1}{4}+x)=0$ or $\frac{4-x}{x+\frac{1}{2}}=0$
$\Rightarrow 4-x+x^2+\frac{1}{a}+x=0$ or $4-x=0$
$\Rightarrow x^2+\frac{5}{4}=0$ or $-x=-4$
$\Rightarrow x=\pm \sqrt{\frac{-5}{2}}$ or $x=4$
But for $x=4$, $\log (4-x)$ is not defined.

And $x=\pm \sqrt{-\frac{5}{2}}=\pm i\sqrt{\frac{5}{2}}$ is complex number, which can't be solution as log is defined for real numbers.
Hence, no solution.