# Why is the exponential function injective but not surjective?

Why is the exponential function injective but not surjective?
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Layne Murillo
The real valued function $f=\mathrm{exp}:\mathbb{R}\to \mathbb{R}$ has the properties that f(0)=1, f′=f and f(x+y)=f(x)f(y) for all $x,y\in \mathbb{R}$. Thus, 1=f(0)=f(x+(−x))=f(x)f(−x) and in particular, $f\left(x\right)\ne 0$ for all $x\in \mathbb{R}$. So, f is not surjective. Since f is continuous, it thus also follows from the intermediate value theorem that f either attains only positive values or only negative values. As f(0)=1, it follows that f(x)>0 for all $x\in \mathbb{R}$. Now, it follows that f′(x)=f(x)>0, and thus f is strictly increasing in $\mathbb{R}$. Every strictly increasing function is injective, thus f is injective.
Interestingly, the exponential function can be extended to $\mathrm{exp}:\mathbb{C}\to \mathbb{C}$, where the function is no longer injective, and attains all complex values except for the sole exception of 0.