the value of sinx varies regularly.
So, graph needs to be analyzwd in intervals.
first thing is that:  log(x) is defined for x>0.
So, sinx>0.
first take interval : 0<x<=pi/2.
sinx is an increasing function for this period and logx is an increasing function.
So, log(sinx) will also increase.
log1 = 0 so, log (sin(pi2))=0.
log(m) is negative when 0<m<1.
So, break it into the intervals and draw the graph.
Note: keep in mind the domain of sinx.

 domain ki range dost  yaar main bhi is tareeke se kar raha hun but tell why does the graph always lie below x-axsis 3-4 quad maain hi kyun

see, it's simple to analyze.
since, x can't be negative it can't go to 2nd and 3rd quad
because sinx will become negative and log of negative numbers is not defined.

now, as Sinx cannot be greater than 1.
and log(m) is always negative for 0<m<1.
at m=1, log becomes zero.
For more clarification you can ask again and read the properties of log for better understanding.