Question

Solve the equation $\sqrt{\log (-x)}=\log \sqrt{x^2}$ (base is 10)

Answer 1

$\sqrt{\log (-x)}=\log \sqrt{x^2}$
If ${\log }_{a}b$ exist $\Rightarrow b>0$
Here we have, $-x>0$
$x<0$
$\therefore \sqrt{x^2}=-x$
$\therefore \sqrt{\log (-x)}=\log (-x)$
$\log (-x)={\left[\log (-x)\right]}^2$
${\left[\log (-x)\right]}^2-\log (-x)=0$
$\log (-x)[\log (-x)-1]=0$
${\log }_{10}(-x)=0,{\log }_{10}(-x)-1=0$
${10}^{{\log }_{10}(-x)}={10}^0,{\log }_{10}(-x)=1$
$-x=1,{10}^{{\log }_{10}(-x)}={10}^1$
$x=-1,-x=10\Rightarrow x=-10$