Question

$\sqrt{4-x}+\frac{1}{{\log }_{10}x}$, the domain is $xϵ(0,1)-\left(k\right).$, what is k

Answer 1

$\sqrt{4-x}$ is defined if $x\le 4$...(1) Again $\frac{1}{{\log }_{10}x}={\log }_{x}10.$ This is defined if x > 0 and $x\ne 1.$
$\therefore xϵ(0,\infty )-1$ ...(2)
From (1) and (2), we can say that
$xϵ(0,1)\cup (1,4)$
or $xϵ(0,1)-\left(1\right).$