Question

find the range of log (sinx) with base of g.i of[x+1]

Answer 1

x-1>0 and (x-1)!=1

D1: x>1 and x!=2

Also sin x must be positive for all x

sin x>0

2nπ<x<2nπ+π where n is any integer.

For n=0

D2: 0<x<π

So domain will be intesection of D1 and D2.

So Domain for n=0 is

(1,π)-{2}=(1,2)union(2,π)

Range calculation:.

When x belongs to (1,π/2)

(x-1) belongs to (0,0.57)

log a(x) is a decreasing function also sin x is increasing function in (0,π/2).Since a is less than 1,it takes all positive value and will be 0 when x=π/2.

So log (x-1)(sin x) will take value from (infinity 0)

Same way we can show that when 2<x<π

(x-1)>1

Also sin x takes value in (0,1) so clearly it takes all negative values.

So we can say that range is all real numbers R

(this answer for this What is the range of logx−1sin(x)logx−1⁡sin⁡(x)?)
u can solve that easily