# logarithmic function

• Mar 27th 2011, 06:17 PM
fran1942
logarithmic function
Hello, I have the following function:

h(x) - lnx + ln(2-x)

The domain of this function is (0,2).
I understand the rule is that you must find a value of x that results in a sum greater than 0.
However I cannot understand why the domain could not be (negative infinity, 2).

For instance if x was -2, then I would have "...ln(2 - - 2)" equalling 4 ?

Thank you for any insight.
• Mar 27th 2011, 06:23 PM
topsquark
Quote:

Originally Posted by fran1942
Hello, I have the following function:

h(x) - lnx + ln(2-x)

The domain of this function is (0,2).
I understand the rule is that you must find a value of x that results in a sum greater than 0.
However I cannot understand why the domain could not be (negative infinity, 2).

For instance if x was -2, then I would have "...ln(2 - - 2)" equalling 4 ?

Thank you for any insight.

Don't forget the other part of your function. At x = -2:
h(-2) = ln(-2) + ln(2 - -2)

ln(-2) does not exist.

-Dan
• Mar 27th 2011, 06:34 PM
Prove It
Is this supposed to be $\displaystyle \displaystyle h(x) = \ln{(x)} + \ln{(2-x)}$?
• Mar 28th 2011, 02:24 AM
mathfun
lnx exists only when x>0
ln(2-x) exists only when 2-x>0 or x<2
So the domain of the function is A=(0,2)