Is $x$ not actually equal to $e^{\ln(x)}$?

Question

Yeah, this is a silly question, but I can't seem to convince myself that the graph $f(x)=x$ is really equal to the graph of $g(x)=e^{\ln(x)}$. Specifically, doesn't this fail on negative values of $x$? Since $g(x)$ is not defined on negative values of $x$, I don't see how these two could be equal. How could I remedy $g(x)$ without using piecewise functions to make these functions have the same domain?

Answer 1

It depends on whether you allow real or complex values for the functions. As other answers point out, if only real values are allowed then $x$ must be positive. If complex values are allowed, then a real value of $x$ might also be negative, and the identity also holds for non-real values of $x$; this is valid for all branches of the logarithm.

BUT ... it is never possible to define any logarithm of zero, therefore even in the more flexible complex case $x\ne 0$ is required.

Answer 2

It happens that $\log$ is a map from $(0,+\infty)$ onto $\mathbb R$ and that $\exp$ is a map from $\mathbb R$ onto $(0,+\infty)$. So, the equality $\exp\bigl(\log(x)\bigr)=x$ only has to be true (and it is true) when $x\in(0,+\infty)$).

Answer 3

The equality $x=e^{\ln(x)}$ holds indeed only for $x>0$.

The piecewise version could be

• for $x>0$ $g(x)=e^{\ln(x)}$
• for $x<0$ $g(x)=-e^{\ln(|x|)}$
• for $x=0$ $g(x)=0$