How can I prove a function is bijective and get its inverse

Question

I have the following function $$f:\Bbb{R}\rightarrow\Bbb{R^+}$$

which is defined as $$f(x) = 2^x$$

How do I show that this function is a bijection and find its inverse?

Answer 1

First we prove that the function is onto. Let us assume that it is into.
Then there exists some real value $y\in\Bbb R^+$ such that there is no real value of x for the condition $$f(x)=2^x=y$$ Taking $log_2$ on both sides, $$x=\log_2 y$$ Since $y>0$, there is a real, defined value of $x$. This is a contradiction. Thus our assumption is incorrect and the function is onto.

Now we prove that the function is one-one. Let us assume that it is many-one.
Then we can take two real values $x_1$, $x_2$ $\in\Bbb R$ such that $x_1\neq x_2$ and $f(x_1)=f(x_2)$, i.e. $$2^{x_1}=2^{x_2}$$ $$2^{x_1-x_2}=1$$ $$x_1-x_2=0$$ $$x_1=x_2$$ This is a contradiction. Thus our assumption is incorrect and the function is one-one.
Since the function is both one-one and onto, it is a bijective function. thus we can find its inverse. $$$$ To find the inverse, we take $y=f(x)$ and then interchange $x$ and $y$ to get $$x=2^y$$ $$\log_2 x=y$$ Therefore the inverse function of f(x) is $$f(x)=\log_2 x$$