# E^ln(x)=x Is it true always?

1. Jul 4, 2011

### vkash

consider this f(x)=eln(sin(x)) f:R-->R.
Can we write this function defination like this f(x)=x f:R-->R
I think no because if we put x as any negative number in first function(function in first line) then there will no solution exist for this but if we put x as any negative number in second function then their will be a solution.
So does it mean that eln(x)=x Is not true always.
ln represent natural log wih base e. R represent set of all real numbers.

2. Jul 4, 2011

### S_Happens

It's only true for x greater than zero, so the positive real numbers onto the real numbers. In my limited experience I believe it is true for any log base.

3. Jul 4, 2011

### dalcde

If you extend the logarithms to negative numbers (to http://en.wikipedia.org/wiki/Complex_logarithm" [Broken]), you do have
$$e^{\ln x}=x$$

Last edited by a moderator: May 5, 2017
4. Jul 4, 2011

### vkash

thanks for confirming my thought.

I have not read any such thing.

Last edited by a moderator: May 5, 2017
5. Jul 4, 2011

### HallsofIvy

Then I presume you haven't dealt with functions of a complex variable.

For real numbers,
$$e^{ln(x)}= x$$
for any x such that ln(x) is defined- i.e. for positive real numbers.

6. Jul 4, 2011

### spamiam

To avoid getting into complex numbers, can't you just use $x = -e^{\ln(-x)}$ for $x<0$?

Or for any nonzero (real) x, $x = \frac{x}{|x|}e^{ln|x|}$?

Last edited: Jul 4, 2011
7. Jul 4, 2011

### dalcde

I apologize if I have gone too far, but my point is that if you somehow extend the logarithms to allow negative numbers, $e^{\ln x}=x$ is true for all x.