What's under the log must be >0
Therefore,
x/(x-1)>0
Let g(x)=x/(x-1)
Make a sign chart to solve this inequality
color(white)(aaaa)xcolor(white)(aaaa)-oocolor(white)(aaaaa)0color(white)(aaaaaaaa)1color(white)(aaaaaaaa)+oo
color(white)(aaaa)xcolor(white)(aaaaaaaa)-color(white)(aaaaa)+color(white)(aaaaaaaa)+
color(white)(aaaa)x-1color(white)(aaaaa)-color(white)(aaaa)-color(white)(aaaa)||color(white)(aaaa)+
color(white)(aaaa)g(x)color(white)(aaaaaa)+color(white)(aaaa)-color(white)(aaaa)||color(white)(aaaa)+
Therefore,
g(x)>0 when x in (-oo,0) uu(1,+oo)
The domain is x in (-oo,0) uu(1,+oo)
To find the range, let
y=log_3(x/(x-1))
So,
By the definition of the logarithm
x/(x-1)=3^y
x=3^y(x-1)
x3^y-x=3^(y)
x(3^y-1)=3^y
x=3^y/(3^y-1)
The denominator must be !=0
3^y-1!=0
=>, y!=0
The range is y in (-oo,0)uu(0,+oo)
graph{log(x/(x-1)) [-8.89, 8.886, -4.45, 4.44]}
