Thread: How do you gind domain and range of these logarithmic functions?

1. How do you gind domain and range of these logarithmic functions?

l(x)=log base 5 of (sqrt(x+5))

AND

m(x)=log(|x|)

AND

n(x)=log((x+1)/(x-1))

I'd like to find the domain and range WITHOUT needing a graphic calculator. Any help would be appreciated!

Thanks!

2. the base has to be positive and different than 1.
the result has to be positive

3. Logs

Hello s3a
Originally Posted by s3a
l(x)=log base 5 of (sqrt(x+5))
First, note that if $\sqrt{x+5}$ is real $x+5 \ge 0$.

So the domain is $\{x \in \mathbb{R} : x \ge -5\}$

And since $\sqrt{(x+5)}$ can take any positive value, its log (to any base) can take all values, positive, negative or zero. So the range is the whole of $\mathbb{R}$.

Originally Posted by s3a
m(x)=log(|x|)
$|x| \ge 0$ for all $x$. But $\log(|x|)$ doesn't exist for $x = 0$, so the domain is $\{x \in \mathbb{R} : x \neq 0\}$

The range again is the whole of $\mathbb{R}$

Originally Posted by s3a
n(x)=log((x+1)/(x-1))
$\log\frac{x+1}{x-1} =\log(x+1)- \log(x-1)$

Remember that the input to a log must be >0. Can you see what the domain is?

And, provided $y$ can take all positive values, $\log (y)$ can take all real values.

4. Originally Posted by s3a
l(x)=log base 5 of (sqrt(x+5))

AND

m(x)=log(|x|)

AND

n(x)=log((x+1)/(x-1))

I'd like to find the domain and range WITHOUT needing a graphic calculator. Any help would be appreciated!

Thanks!
If $a > 0$ and $a \neq 1$ then the domain of the function $f(x)=\log_a(x)$ is $d = \mathbb{R}^+$ and the range of the function is $r = \mathbb{R}$

to l:

$l(x)=\log_5(\sqrt{x+5})$
Calculating the domain:

$\sqrt{x+5}>0~\implies~x+5>0~\implies~x>-5$

Since $\sqrt{x+5} \in (0, +\infty)$ $r = \mathbb{R}$

to m:

According to my previous considerations you'll get: $d = \mathbb{R} \setminus \{0\}$ and $r = \mathbb{R}$

to n:

$n(x)=\log\left(\dfrac{x+1}{x-1}\right)$
Calculating d:

$\dfrac{x+1}{x-1}>0~\implies~\left(x+1>0\wedge x-1>0 \right)~\vee~\left( x+1<0\wedge x-1<0 \right)$ and therefore:

$d=(-\infty,-1)\cup(1,+\infty)$

If $x < -1$ then $0 < \dfrac{x+1}{x-1} <1$ and therefore $n(x) < 0$;

If $x > 1$ then $1 < \dfrac{x+1}{x-1}$ and therefore $n(x) > 0$;

Since $\lim_{|x|\to \infty}\left( \dfrac{x+1}{x-1} \right) = 1$ the graph of n approaches the x-axis.

Therefore $r = \mathbb{R}\setminus \{ 0\}$