How do we 'know' that $2^x$ is continuous?

Question

It is intuitive for $2^n$, if $n$ is an integer, to exist.

How do we know that less intuitive values such as $2^\frac{1}{2}$, $2^\sqrt{2}$, $2^\pi$ etc exist?

I'd like to accept that $2^x$ is continuous, but how can we be sure of the existence of the number when $x$ is something obscure, like an irrational number?

Answer 1

This is usually dealt with by defining $$2^x = e^{x \ln 2}$$ Such a definition works provided we ultimately have a way to define $e^x$ so we can be sure it's continuous.

One way this is done is by defining the exponential function as the the inverse of natural logarithm, which is defined as the definite integral $$\ln x = \int_1^x \frac{dt}{t}$$ This integral is known to exist and be continuous by theorems about integrals of continuous functions on closed intervals. As $\ln$ is also strictly monotonically increasing, the inverse exists and is also continuous

Answer 2

Read this answer. It's pretty good. We just can define these numbers in "most intuitive" way.