I want to find a function that satisfies

$$\Delta [f(x)]=f'[x]$$

Obviously the solution is the exponential function $f(x)=a^x$ with $a$ in between $2$ and $e$ because $\Delta[2^x]=2^x$ and $(e^x)'=e^x$.

Thus the base should satisfy the equation

$$\frac{a^x}{a-1}=a^x \log a$$

or after contraction,

$$\frac{1}{a-1}=\log a$$

This is equal to

$$a^a-ea=0$$

I wonder what is the solution?