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

1 Answer
Dec 25, 2017

Answer:

For domain, the argument of the logarithm must be greater than 0.
Domain: #\{x | x > 6, x \in \mathbb{R}}#
Range: #\{y | y \in \mathbb{R}}#

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=\log(2x-12)#
#y = \log_{10}(2x-12)#

This can be represented by, in exponential form,

#10^y = 2x - 12#

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

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

#2x - 12 > 0#
#\ \ \ \ \ \ \ \ \ 2x > 12#
#\ \ \ \ \ \ \ \ \ \ \ x > 6#

The domain is #\{x | x > 6, x \in \mathbb{R}}#

For range

For any logarithmic function of the form #y = a \log_c(b(x-h))+k#, the range is #\{y | y \in \mathbb{R}}#.