We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \frac{1}{\log{\left(x \right)}} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \frac{1}{\log{\left(\sin{\left(x \right)} \right)}} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\log{\left(\sin{\left(x \right)} \right)}}{\log{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \frac{1}{\log{\left(x \right)}}}{\frac{d}{d x} \frac{1}{\log{\left(\sin{\left(x \right)} \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\log{\left(\sin{\left(x \right)} \right)}^{2} \sin{\left(x \right)}}{x \log{\left(x \right)}^{2} \cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\log{\left(\sin{\left(x \right)} \right)}^{2} \sin{\left(x \right)}}{x \log{\left(x \right)}^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\log{\left(\sin{\left(x \right)} \right)}^{2} \sin{\left(x \right)}}{x \log{\left(x \right)}^{2}}\right)$$
=
$$1$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)