how to prove the uniqueness and existence of equations I've the equation $e^x=5$, know it has the solution $x=\ln 5$. How to prove the existence before, and after the uniqueness of this solution?
 A: Let $f(x)=e^{x}$. f is continuous on $\mathbf{R}$ and strictly increasing, from 0 at $-\infty$ to $\infty$ at $\infty$, hence by intermediate value theorem ( a stronger version for increasing functions) $f(x)=k$ where $k>0$ has a unique solution!
A: It's sort of circular if you use the expression $\ln 5$, because this symbol is only defined thanks to existence and uniqueness for all equations $$e^x=k$$
for $k>0$.
So let's rewind a bit and avoid the circularity, by not mentioning the logarithm. You could do an existence proof, for example, by using Bolzano's theorem on the function $e^x -5$, for example. Whatever your definition of $e^x$ is, to do what I've mentioned all you need are two things: $$\begin{align}&\text{1) }2\leq e^1 \leq 3 \\ &\text{2) } e^n = e^1\stackrel{(n)}{\cdots}e^1\, \forall n \in\Bbb N\end{align}$$
(and that the exponential function is continuous)
Then, to prove uniqueness you simply prove the injectivity of the exponential function, for example by proving that it's strictly increasing. Again, regardless of the definition you have for the exponential, we can do this using two properties: $$\begin{align}&\text{1) } e^x >0 \, \forall x\in\Bbb R \\ &\text{2) } e^{x+y} = e^xe^y\end{align}$$
All of this should suffice for a proof of what you want, and can be easily extended to any such equation (taking $y\in \Bbb R $ instead of 5), and thus allow you to define the logarithm. This is however, far, far, far from the only way! Also, of course, the proof of the properties I mention will heavily depend on how you define $e^x$, so they're up to you to prove, if you wish.
