How do you find the domain and range of #h(x) = log_3[x/(x - 1)]#?

1 Answer
Jul 25, 2018

Answer:

The domain is #x in (-oo,0) uu(1,+oo)#. The range is #y in (-oo,0)uu(0,+oo)#

Explanation:

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)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##0##color(white)(aaaaaaaa)##1##color(white)(aaaaaaaa)##+oo#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaaaaaa)##+#

#color(white)(aaaa)##x-1##color(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]}