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

Jul 25, 2018

The domain is $x \in \left(- \infty , 0\right) \cup \left(1 , + \infty\right)$. The range is $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

#### Explanation:

What's under the $\log$ must be $> 0$

Therefore,

$\frac{x}{x - 1} > 0$

Let $g \left(x\right) = \frac{x}{x - 1}$

Make a sign chart to solve this inequality

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a a}$$1$$\textcolor{w h i t e}{a a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$g \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$g \left(x\right) > 0$ when $x \in \left(- \infty , 0\right) \cup \left(1 , + \infty\right)$

The domain is $x \in \left(- \infty , 0\right) \cup \left(1 , + \infty\right)$

To find the range, let

$y = {\log}_{3} \left(\frac{x}{x - 1}\right)$

So,

By the definition of the logarithm

$\frac{x}{x - 1} = {3}^{y}$

$x = {3}^{y} \left(x - 1\right)$

$x {3}^{y} - x = {3}^{y}$

$x \left({3}^{y} - 1\right) = {3}^{y}$

$x = {3}^{y} / \left({3}^{y} - 1\right)$

The denominator must be $\ne 0$

${3}^{y} - 1 \ne 0$

$\implies$, $y \ne 0$

The range is $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

graph{log(x/(x-1)) [-8.89, 8.886, -4.45, 4.44]}