How to Find the number of tangents to the curve y=$f\left(x\right)$ parallel to line $x+y=0$ 
Question For $x$$>0,$ let
$f\left(x\right)=\int_{1}^{x}\left(\sqrt{\log t}-\frac{1}{2}\log\sqrt{t}\right)dt$
The number of tangents to the curve y=$f\left(x\right)$parallel to
  line x+y=0 is _________________


MY approach $x+y=0\Longrightarrow\frac{dy}{dx}=-1$
$f\left(x\right)=\int_{1}^{x}\left(\sqrt{\log t}-\frac{1}{2}\log\sqrt{t}\right)dt$$\Longrightarrow f'\left(x\right)=$$\sqrt{\log x}-\frac{1}{2}\log\sqrt{x}$
I know that parallel lines have same slope,but it would not give me
number of tangents.

Hint: Finding the number of tangents of $f(x)$ parallel to line $y=-x$ is equivalent to finding the number of roots of the equation $f'(x)=-1$.
