How do you find the domain of $f (x) = \frac{log [x - 4]}{log [3]}$ $f(x)=(log[x-4])/(log[3])$ ?

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How do you find the domain of $f (x) = \frac{log [x - 4]}{log [3]}$ $f(x)=(log[x-4])/(log[3])$ ?

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Answer 1 · 2015-05-21T22:29:16

The domain of function $f (x)$ $f(x)$ is the interval $(4; + \infty)$ $(4;+infty)$ .

First you need to know that you can evaluate $log (x)$ $log(x)$ only when $x$ $x$ is positive ( $x > 0$ $x>0$ ) and equals $0$ $0$ only when $x = 1$ $x=1$ .

So, in our denominator we don't any problems, $3$ $3$ is a positive number different that $1$ $1$ so we get a nonzero denominator.

Our numerator consists of $log (x - 4)$ $log(x-4)$ , which can be $0$ $0$ but this doesn't concern us particularly. The number inside, $x - 4$ $x-4$ , must be positive, though. So:

$x - 4 > 0$ $x-4>0$
$x > 4$ $x>4$
$x \in (4; + \infty)$ $x in (4;+infty)$

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How do you find the domain of $f (x) = \frac{log [x - 4]}{log [3]}$ $f(x)=(log[x-4])/(log[3])$ ?

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How do you find the domain of $f (x) = \frac{log [x - 4]}{log [3]}$ $f(x)=(log[x-4])/(log[3])$ ?