Why is the exponential function injective but not surjective?

Taniya Melton 2022-10-15 Answered
Why is the exponential function injective but not surjective?
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Answers (1)

Layne Murillo
Answered 2022-10-16 Author has 14 answers
The real valued function f = exp : R R has the properties that f(0)=1, f′=f and f(x+y)=f(x)f(y) for all x , y R . Thus, 1=f(0)=f(x+(−x))=f(x)f(−x) and in particular, f ( x ) 0 for all x 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 R . Now, it follows that f′(x)=f(x)>0, and thus f is strictly increasing in R . Every strictly increasing function is injective, thus f is injective.
Interestingly, the exponential function can be extended to exp : C C , where the function is no longer injective, and attains all complex values except for the sole exception of 0.
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