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There are a variety of different ways to define the natural exponential function, but the definition that feels the most conceptually elegant to me is to define it as the function that is its own derivative, and that $\text{exp}(0)=1$. I am currently trying to figure out if it is possible to derive all the properties of $e^x$/$\text{exp}(x)$ merely from this fact. In addition, I have defined the natural logarithm to simply be the inverse of $\text{exp}(x)$, which was enough to determine that the derivative of $\ln{x}$ is $\frac{1}{x}$. I have been trying in vain to prove from this definition that $\forall{k}\in\mathbb{R},\text{exp}(x+k)=\text{exp}(x)\space\cdot\space\text{exp}(k)$.

Null Spark
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    FYI, there's more than just $y = e^x$ and $0$, there's any function of the form $y = Ae^x$ for some constant $A$. If you want to uniquely define the exponential function from a differential equation, it's customary to include the initial value condition $y(0) = 1$. – user837206 Oct 30 '20 at 14:53
  • @user837206 Ahh, thank you for the correction. I did forget to include the initial condition. I will edit the post to include that. – Null Spark Oct 30 '20 at 14:54

2 Answers2

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Note that (provided $\exp(k)\ne0$), we have for $f(x):=\frac{\exp(x+k)}{\exp(k)}$, $$f(0)=\frac{\exp(0+k)}{\exp(k)}=1,\qquad f'(x)=\frac{\exp'(x+k)}{\exp(k)}=\frac{\exp(x+k)}{\exp(k)}=f(x)$$ so that by uniqueness of the solution, $f(x)=\exp(x)$ for all $x$. Hence, $$ \exp(x+k)=\exp(x)\exp(k)\quad \text{if }\exp(k)\ne0.$$ As $\exp(0)=1$, we have $\exp(k)\ne0$ in a neighbourhood of $0$, hence if $\exp(x)=0$, then $\exp$ is also $=0$ in a neighbourhood of $x$. It follows that $$\{\,x\in\Bbb R\mid \exp(x)=0\,\} $$ is open - but it trivially is also closed, and is non-empty. This means that it is all of $\Bbb R$. Hence we can strike our caveats and have $$ \exp(x+k)=\exp(x)\exp(k)\quad \text{for all }x,k\in \Bbb R.$$

Hagen von Eitzen
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Both expressions solve the initial value problem $$y'(x)=y(x) ,\quad y(0)=\exp(k),$$ which has a unique solution. So they must be equal.

(More conceptually, that identity is a reflection of the fact that solutions of first-order ODEs are flows.)

Hans Lundmark
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