is $f(x) = 2^x$ bijective?

Question

I've been posed with the question "Why is $f$ not invertible?"

I have learned that $x^2$ is not bijective unless I restrict it to only use non-negative Reals. However when I look at the curve of $2^x$ it looks to me that it passes the 1 unique x point throughout the whole chart. However, I need to prove why $2^x$ is not invertible and rewrite it to make it invertible. Do I need to restrict it from $R \rightarrow R^+$ ? What else am I not understanding? Thank you.

Answer 1

Note that the function $f(x)=2^x$ is not surjective because: $$\Im(f(x))=R^+$$ Being not surjective, $f(x)$ is not bijective. In this case we can't find the inverse because the inverse exists only when $f(x)$ is bijective.

If you restrict the codomain on $R^+$, the function is injective and surjective, so it's bijective. As noted before, the inverse is: $$y=g(x)=\log_2(x)$$

Answer 2

Depend from where to where. But indeed, it's at least injective on $\mathbb R$, so on some co-domain, it will obviouly be bijective. Notice that whenever $a>0$, then $a^x$ is nothing else than $e^{x\ln(a)}$.