# Simple logarithm simplifications

1. Jul 28, 2009

### ehilge

1. The problem statement, all variables and given/known data
Hey all, I just need some help remembering my basic simplification rules.

Can the expression eC*ln(x) be simplified anymore. It would certainly help out an integral I'm working on but I don't remember my simplification real well.

2. Relevant equations
elnx=x

3. The attempt at a solution
I'm doubting there is a solution since eC*ln(x) = eclnx and I don't see how that helps much. But I thought I'd check with you guys and see if anyone had a better idea.
Thanks!

2. Jul 28, 2009

### HallsofIvy

No, $e^{C ln(x)}$ is NOT equal to $e^{C^{ln x}}$. I don't know where you got that. It is equal to $e^{ln(x^C)}$ (because [/itex]C ln(x)= ln(x^C)[/itex]).

Last edited by a moderator: Jul 28, 2009
3. Jul 28, 2009

### ehilge

ohhhhh, I see now, thanks!

edit: also ecln(x) is equivalent to ecln(x)
this is the same as saying that 23*2 = 232 = 64

Last edited: Jul 28, 2009
4. Jul 28, 2009

### Kaimyn

Are you sure? Because 23*2=26 and 232=29. What you said is equivalent to saying 4*6 = 46 or 64.

As HallsofIvy said, ecln(x)=eln(xc). This can be simplified. Can you see how?

5. Jul 28, 2009

### HallsofIvy

No, it is not. You are not distinguishing between
$$\left(e^a\right)^b$$
and
$$e^{(a^b)}$$

What you are writing, eab, is equivalent to
$$e^{\left(a^b\right)}$$
which is NOT the same as
$$\left(e^a\right)^b= e^{ab}$$.

6. Jul 28, 2009

### ehilge

ya, you're right, I meant to write what you have in the 2nd part there, I just wasn't liberal enough with parenthesis. Sorry, my bad.

so what I should have written in the first place is ecln(x) = (ec)ln(x) , right?

also, Kaimyn, eln(xc) simplifies to xc, if I remember my log stuff right (which obviously isn't a given) the e and the ln should essentially cancel each other out.

thanks

7. Jul 28, 2009

### HallsofIvy

Yes, ec ln(x)= eln xc= xc.

8. Jul 31, 2009

### Дьявол

He thought:

$$(e^c)^{ln(x)}=e^{c*ln(x)}=e^{ln(x^c)}$$