Are real numbers enough to solve simpler exponential equations such as $2^x=5$, $(1/e)^x=3$, and $\pi^x=e$?

Question

How to prove that solutions of simpler exponential equations (*) are real numbers?

In other words, how to prove that set of real numbers is enough to solve something like $2^x = 5,$ or $(\frac{1}{e})^x = 3$ or even $\pi^{x} = e$?

(*) Assumption is that base is not something like $e^{\sqrt{2}}, e^{\pi}, 2^{\frac{1}{2}},$ (#) so only "given" (computed in some way) indeed real? Also, inverse question: How to prove that this numbers (#) are real?

Graphically: how to prove that exponential function $f(x) = e^x$ for $x \in (- \infty, + \infty)$ "picks up" all the real values?

Answer 1

First recall that the map \begin{align*} \exp &: \mathbb{R} \to (0, \infty) \end{align*} is a bijection (you can prove this by e.g., proving that $\exp$ is continuous, strictly increasing (this gives injective), unbounded, and obtains arbitrarily small positive values - then invoke the intermediate value theorem (this gives surjective)). In particular $\exp(x) = r$ has a unique real solution whenever $r \in \mathbb{R}_{> 0}$, which we call $\log(r)$.

By definition, whenever $\alpha \in \mathbb{R}_{>0}$ we have $$\alpha^x = \exp (\log(\alpha) x)$$

which defines a bijection $\mathbb{R} \to (0, \infty)$ whenever $\log(\alpha) \neq 0$ - i.e., when $\alpha \neq 1$ (since then it is a composition of bijections).

This also gives a solution to your second question, we know that $\exp(x) \in \mathbb{R}$ whenever $x \in \mathbb{R}$. Since $\log(\alpha) \in \mathbb{R}$ exists whenever $\alpha \in \mathbb{R}_{>0}$ we have $\alpha^x = \exp(\log(\alpha)x) \in \mathbb{R}$ whenever $x \in \mathbb{R}$.

Note that this question is much more subtle when $\alpha < 0$ since we must choose our $\log$'s etc. But for sure $(-1)^{1/2} \notin \mathbb{R}$. Moreover, your equations will also have complex solutions - so we do not have uniqueness if we extend our view to $\mathbb{C}$. The moral being - $\log$ is not well behaved in $\mathbb{C}$ so your question turns out to be a subtle one if we allow $\mathbb{C}$ to be involved.

Answer 2

Some of the equations you mention actually have complex solutions as well.

But in general, you may ask - given an equation, if I solve it using real methods - how do I know there are no other complex solutions? The answer is you cannot. Real numbers are a subset of the complex numbers, so you are actually working within a small subset of ${\mathbb{C}}$. You can, however, redefine your variable to be a complex variable and solve it using complex methods - and if you still end up with purely real solutions (that is, solutions with imaginary part ${=0}$), you can be sure those are indeed the only solutions (at least within the context of ${\mathbb{C}}$, anyways).

Now, asking how to tell if expressions like ${2^e}$ are real completely depends upon the definitions you use for these expressions. One definition for ${2^e}$ is to take a limit as ${n\rightarrow\infty}$ of ${2^{e_n}}$ - where ${e_n}$ is a sequence of real numbers converging to $e$, taking the principle root at each point in the sequence (by principle root, I mean for example taking ${2^{\frac{1}{2}}}$ as being ${\sqrt{2}}$). In this case, the expression, by definition, will intrinsically be real.