# How do you find the domain and range of y = log(2x -12)?

Dec 25, 2017

For domain, the argument of the logarithm must be greater than 0.
Domain: $\setminus \left\{x | x > 6 , x \setminus \in \setminus m a t h \boldsymbol{R}\right\}$
Range: $\setminus \left\{y | y \setminus \in \setminus m a t h \boldsymbol{R}\right\}$

#### Explanation:

For domain:

The argument of the logarithm (stuff inside the log) must be greater than 0. This is because logarithm can be viewed as the inverse of an exponential function. For example:

$y = \setminus \log \left(2 x - 12\right)$
$y = \setminus {\log}_{10} \left(2 x - 12\right)$

This can be represented by, in exponential form,

${10}^{y} = 2 x - 12$

10 raised to any exponent cannot get a negative number or be equal to zero, thus $2 x - 12 > 0$

So solve $2 x - 12 > 0$ to get the permissible values for $x$.

$2 x - 12 > 0$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus 2 x > 12$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus x > 6$

The domain is $\setminus \left\{x | x > 6 , x \setminus \in \setminus m a t h \boldsymbol{R}\right\}$

For range

For any logarithmic function of the form $y = a \setminus {\log}_{c} \left(b \left(x - h\right)\right) + k$, the range is $\setminus \left\{y | y \setminus \in \setminus m a t h \boldsymbol{R}\right\}$.