Question about why ln(e^x) =/= x

[physicsernaw]

Why is ln(e^x) =/= x?

The domain and range of the LHS are the same as the RHS, so I don't understand why this equation is false, where e^ln(x) = x, and the LHS and RHS of this does not have the same domain...

I know that e^x and ln(x) are inverse functions, so please don't only tell me this. Why does e^ln(x) = x, while ln(e^x) =/= x?

EDIT:
Like, I understand why sin^-1(sin(x)) =/= x, whereas sin(sin^-1(x)) = x, and this is because sin^-1(x) has a range of -pi/2 to pi/2

 

[Vorde]

They are equal.

ln(e^x) = x*ln(e) = x

What/who told you differently?

 

[physicsernaw]

They are equal.

ln(e^x) = x*ln(e) = x

What/who told you differently?

I could have sworn wolfram alpha was telling me the equation is false, but I just tried again and it told me the equation is indeed true

 

[Mark44]

It's the other way around that you run into domain considerations:

$$ e^{\ln x} = x$$

The left side is defined only for x > 0 (considering only the real-valued log function).