# Domain Logarithmic Function (Unsolved)

• April 6th 2008, 07:47 PM
wickwiki
Domain Logarithmic Function (Solved)
f(x)=-2ln(x+1)

Find the domain of f.

So, far this is I have:
f(x)=-2ln(x+1)
f(x)=-2e^(x+1)

Can anyone go over this problem step by step with me?
• April 6th 2008, 07:53 PM
Mathstud28
Ok to find the domain
for $f(x)=-2\ln(x+1)$...the only problem is that $x+1\geq{0}$ due to $\ln(x)$s properties...so $x+1\geq{0},x\geq{-1}$
• April 6th 2008, 07:55 PM
TheEmptySet
Quote:

Originally Posted by wickwiki
f(x)=-2ln(x+1)

Find the domain of f.

So, far this is I have:
f(x)=-2ln(x+1)
f(x)=-2e^(x+1)

Can anyone go over this problem step by step with me?

The domain of

$g(x)=\ln(x)$

is the set $(0,\infty)$

so your function is related to the unshifted natrual log by

$f(x)=-2g(x+1)=-2\ln(x+1)$ This shifts the graph one unit left in the x direction so the domain is

$(-1,\infty)$
• April 6th 2008, 07:56 PM
Jhevon
Quote:

Originally Posted by Mathstud28
for $f(x)=-2\ln(x+1)$...the only problem is that $x+1\geq{0}$ due to $\ln(x)$s properties...so $x+1\geq{0},x\geq{-1}$

it must be strictly greater than zero. log(0) is undefined as well
• April 6th 2008, 08:03 PM
Mathstud28
O man
I had it that way...but I was thinking of $\sqrt{x}$ and I went back and changed it...on these forums I try to answer as fast as I can so I get messed up sometimes...thanks!...I knew that by the way =D
• April 6th 2008, 08:13 PM
Jhevon
Quote:

Originally Posted by Mathstud28
...I knew that by the way =D

that's ok

yes, i know you knew :)
• April 6th 2008, 08:15 PM
Mathstud28
Haha
::blushes:: haha thanks
• April 6th 2008, 08:24 PM
wickwiki
OK, solution solved thanks for the help.

I would like to know a bit more about how you solved it.

f(x)=-2ln(x+1)
So, what happens with the negative two exactly?

Because the one of my similar problems I solved ln(x+4) and I got
Domain= (-4,\infty).

So, how is the -2 effect the equation?
• April 6th 2008, 08:27 PM
Mathstud28
It doesnt
The only thing that matters in a $\ln(u(x))$ is that the domain is restricted so that $\forall{x}\in{Domain},u(x)>0$...make sense?