Simple logarithm simplifications

[ehilge]

Homework Statement

Hey all, I just need some help remembering my basic simplification rules.

Can the expression e^C*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.

Homework Equations

e^lnx=x

The Attempt at a Solution

I'm doubting there is a solution since e^C*ln(x) = e^clnx 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!

 

[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]).

 

[ehilge]

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]).

ohhhhh, I see now, thanks!

edit: also e^cln(x) is equivalent to e^cln(x)
this is the same as saying that 2^3*2 = 2³² = 64

 

[Kaimyn]

ohhhhh, I see now, thanks!

edit: also e^cln(x) is equivalent to e^cln(x)
this is the same as saying that 2^3*2 = 2³² = 64

Are you sure? Because 2^3*2=2⁶ and 2³²=2⁹. What you said is equivalent to saying 4*6 = 4⁶ or 6⁴.

As HallsofIvy said, e^cln(x)=e^ln(xc). This can be simplified. Can you see how?

 

[HallsofIvy]

ohhhhh, I see now, thanks!

edit: also e^cln(x) is equivalent to e^cln(x)
this is the same as saying that 2^3* = 2³² = 64

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

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

 

[ehilge]

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

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 e^cln(x) = (e^c)^ln(x) , right?

also, Kaimyn, e^ln(xc) simplifies to x^c, 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

 

[HallsofIvy]

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 e^cln(x) = (e^c)^ln(x) , right?

also, Kaimyn, e^ln(xc) simplifies to x^c, 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

Yes, e^{c ln(x)}= e^{ln xc}= x^c.

 

[Дьявол]

He thought:

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