# Absolute value question

1. Aug 12, 2005

### asdf1

if h=ln(absolute x)
then how do you calculate e^h?

2. Aug 12, 2005

### TD

If $h = \ln \left| x \right|$ then $e^h = e^{\ln \left| x \right|} = \left| x \right|$.

Seems simple, do you mean this?

3. Aug 13, 2005

### asdf1

that's what i originally thought,
but the correct answer doesn't need absolute value...
i think that's strange...

4. Aug 13, 2005

### VietDao29

It must be |x|. Have you ever seen:
$$e ^ {\mbox{something}} = \mbox{a negative number}$$??
Viet Dao,

5. Aug 13, 2005

### lurflurf

yes

$$e^{\pi i}=-1$$

6. Aug 13, 2005

### rachmaninoff

Which brings up an important point, $e^{ \ln x }$ depends on whether we're working in the Reals or the Complexes. The real logarithm is undefined for x < 0.

7. Aug 13, 2005

### VietDao29

Okay, my something is $$\in \mathbb{R}$$, not C .
Viet Dao,

8. Aug 13, 2005

### lurflurf

What were the preceding steps? This type of thing happens alot when what you were finding was the absolute value of something and droping it gives the something.
for instance if
xy'=y y(0)=0 (the solution being y=x or y=-x)
one might get
|y|=|x| as an answer but though error have
y=|x|

9. Aug 13, 2005

### TD

Well, you can always take a numerical example if you're in doubt...:

$$\begin{array}{l} e^{\ln \left( { - 1} \right)} = - 1 \\ e^{\ln \left| { - 1} \right|} = 1 \\ \end{array}$$

10. Aug 13, 2005

### master_coda

This is incorrect. Assuming we're working in the reals $e^{\ln(-1)}$ isn't defined, because $\ln$ is undefined for negative numbers and so any expression including it is also undefined.

11. Aug 13, 2005

### TD

My example wasn't necessarily over the reals, I don't see why we'd have to exclude to complex numbers - at least not for this example which is only intended to show that abs values are needed.

12. Aug 13, 2005

### lurflurf

Again where how did this arise. One way absolute values often arise in arguments of logarithums is when one wants an antiderivative of a logarothmic derivative that is restricted to reals and valid almost every where.
ie
xy'=1 y(e)=1
y=log(|x|)
clearly if we are working with complex numbers
y=log(x) would do
and sometimes errors arise as when
xy'=y y(1)=1
clearly
y=x
when restricted to real one often finds
|y|=exp(log(|x|))=|x|
but a common error is to get
y=|x| which satisfys the differential equation every where except the origin
if you log(|x|) did not arise though integrating a logarithmic derivative perhaps some other error was made to cause you to get |h| when h was wanted (or mayber h>0 so it does not matter).

13. Aug 14, 2005

### asdf1

Originally, I was trying to solve this problem:
dy/dx + ytanx = sinx, y(0)=1

The first step I did was to integrate tanx and then there came the absolute value dilema...

14. Aug 14, 2005

### lurflurf

So it was due to integrating a logarimic derivative.
Say you were to solve th associated homogeneous problem in preparation for variation of parameters.
y'+y tan(x)=0
y'/y=cos'(x)/cos(x)
log(|y|)=a+log(|cos(x)|)
|y|=|C||cos(x)| then let C absorb the sign
y=C*cos(x)
which is clearly a solution
of course if we were working with complex numbers we would never have introduced the absolute value. This problem arises though finding the function. It has nothing to do with the function itself.

15. Aug 14, 2005

### HallsofIvy

Staff Emeritus
Because the original poster said his question only concerned real numbers and to go outside that without specifically saying so can be confusing.

16. Aug 14, 2005

### TD

Sorry, my bad then although I nowhere saw asdf1 posting before me that it concerned the reals, only Vietdao.

17. Aug 14, 2005

### VietDao29

So isn't it obvious that:
$$\ln|x| \in \mathbb{R} \mbox{, \forall x \in \mathbb{R} - \{0\}}$$?????
Viet Dao,

18. Aug 15, 2005

### asdf1

so llurflurf you mean that there is supposed to be an absolute value sign, except that it's already included in "c"?

19. Aug 22, 2005

### asdf1

the constant in this question doesn't seem to absorb...

that question seems to work, but in this question:
y`= -y/x, y(1)=1
the same problem arises, but the constant doesn't absorb the negative sign...
it's weird~

20. Aug 22, 2005

### lurflurf

What do you mean
y'=-y/x y(1)=1
y'/y=-1/x
log(|y|)=-log(|x|)+a=log(1/|x|)+a
|y|=exp(a)/|x|
|y|=|c|/|x|
y=c/x
y(1)=c/1=c=1
y=1/x