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

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?

• Because every value is smooth. There isn't a positive real that will make it jump suddenly. – Calculus Nov 9 '14 at 12:07
• Your question and your question's title don't match: you don't want to know about continuity of $\;2^x\;$ but rather about how is that defined when $\;x\;$ is not "a nice" real number. – Timbuc Nov 9 '14 at 12:07
• The title is matching my question. My question is regarding the continuity of $2^x$. But to be sure of the continuity, we must be sure that the curve doesn't have any 'holes' in it, hence why I am questioning the existence of a function value when $x$ is 'not nice'. – Trogdor Nov 9 '14 at 12:09

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