Question

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

Answer 1

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]}