Closed form solution to $\frac{1}{a-1}= \log a$

Question

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?

Answer 1

The solutions to this equation cannot even be expressed in terms of Lambert's W function, let alone in terms of elementary ones. Their numerical values, however, are

$${\large a}_1=0.2592463566483049762658328267922644867392517823397669639360784269929\ldots$$

$${\large a}_2=2.2399778876565500700638977660928972654111884831688969821454636850130\ldots$$