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

[vkash]

consider this f(x)=e^ln(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 e^ln(x)=x Is not true[/color] always.
ln represent natural log wih base e. R represent set of all real numbers.

 

[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.

 

[dalcde]

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

 

[vkash]

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.

thanks for confirming my thought.

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

I have not read any such thing.

 

[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.

 

[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, $x = \frac{x}{|x|}e^{ln|x|}$?

 

[dalcde]

I have not read any such thing.

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.