Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need help finding the domain and range of logarithm functions. For example, what is the domain and range of $y=\log(x-3)$?

share|improve this question
1  
If $y=\log_ax$ where real $a>0;$ we have $x=a^y>0\implies x$ must be $x>0$ –  lab bhattacharjee Aug 11 '13 at 18:03
3  
Ok, why does this have $14197$ views? –  LTS May 18 at 15:26
1  
@Oliver From google and other search engines. Someone gets a homework problem and wants to figure out how to "find the domain and range of a logarithmic function" and boom, this pops up. –  Michael T May 18 at 15:42
add comment

4 Answers 4

up vote 6 down vote accepted

Let's take a look at the function $$y = \log(x - 3)$$

We are trying to find the Domain and the Range of this function, recalling that:

  • Domain: Includes all values of $x$ for which the function is defined.
  • Range: Includes all values $y$ for which there is some $x$ such that $y = \log(x - 3)$.

    It makes no sense to write $y = \log(a)$ when $a \leq 0$ because $\log(a)$ is defined only for positive $a$. So in this problem, $y = \log(x = 3)$, is defined if and only if $x - 3 \gt 0 \iff x \gt 3$, and that gives you the domain $x\in (3, +\infty)$.

The range of $y$ is all of $\mathbb R$.

Graphing a function is a great way to confirm or gain insight into the domain and range of a function.

E.g. Below is the graph of the function $y = \log_e(x - 3)$.

enter image description here

As $x$ grows in size, so to does $y$. That is, $y$ is an increasing function. As $x$ approaches $3$ from the right, although $y=\log_e(x - 3)$ is not defined for any $x \leq 3$, within the interval $x \in (3, 4)$, $y < 0$. The closer $x$ is to $3$ (from the right), the smaller $y$ gets (the "more negative" $y$ becomes.)

share|improve this answer
    
I sometimes add 1+1 incorrectly! :-) –  Amzoti Aug 12 '13 at 0:19
    
@amWhy: but i add 1+1 correctly. :-) –  B. S. Aug 12 '13 at 6:00
add comment

Remember that for $x \in \mathbb{R}$ the argument of $\log$ cannot be negative and $\log(x)$ varies from $-\infty$ to $\infty$

share|improve this answer
add comment

$\log$ is the inverse function of exponentiation: $y = \log_b x \iff x = b^y$ (only for positive $b$). So the domain of $y = \log_b x$ is the range of $x = b^y$ which is all $x$ with $x > 0$ since $b^y$ is always positive for positive $b$. And the range of $y = \log_b x$ is the domain of $x = b^y$ which is any number $y \in \mathbb{R}$.

Now for other variations, just apply the above. In your example we have $y = \log(x - 3)$. the domain of $\log$ is positive numbers and so we must have $x - 3 > 0 \implies x > 3$. The range is still all real numbers.

share|improve this answer
add comment

$y = \log(x-3)$

$x-3>0$

$x>3$

so domain is $(3,\infty)$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.