Is it possible to rewrite $exp(x)$ with $x$ as the base

Question

Simple question, rewriting $e^x$ with $x$ as the base- or more generally, trying to remove $x$ from the exponent. I don't know if it's possible, as I played around with it for a little while and got nothing. If it is, I suspect it has something to do with the Lambert W function. If it's not possible, is there an intuitive reason as to why not or is it just something that can't be done algebraically?

I'm using it to simplify a string of terms into a shorter string of terms with $x$ as the base for each one. It's just convenient if the answer could be given in this polynomial-like form.

Thanks, and sorry if I was a bit vague or missed details.

Edit: It really is what's on the label; $e^x = x^?$- it's meant to be general.

If you're solving for $?$ in the situation above,

$e^x=x^y$

$y=\log_x(e^x)$

$y=x*log_x(e)$

$y=\frac{1}{\log_{e^x}(x)}$

And from there, I don't know how to go about reducing the equation to a single $x$, let alone removing the log and putting it in the form of $x^y$. Any ideas?

Answer 1

There's no $y$ such that $e^x = x^y$ for all $x$, because in the long term, the exponential beats any polynomial.

One way to see it, is when defining the exponential as a power series: $e^x := \displaystyle\sum_{n \geq 0} \dfrac{x^n}{n!}$. In particular, if the expression you're trying to simplify has $x \approx 0$, you can use the expression above and neglect terms above a certain order (that's called a MacLaurin/Taylor series). For example, an approximation of order $2$ would be $e^x = 1 + x + \dfrac{x^2}{2} + O(x^3)$, where $O(x^3)$ denotes the remainder, which is of order $x^3$.

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Another expression, which is not in polynomial form but might come in useful, is Euler's relation $e^{ix} = \cos x + i\sin x$.

Answer 2

$$e^0=1\text{ vs. }0^y=0$$ precludes any solution of the form $$e^x=x^y.$$

(And as shown by @nala, no polynomial can do. We can also add that no fractional polynomial can do. Furthermore, it is know that $e^x=1+x+\frac{x^2}2+\frac{x^2}{3!}+\cdots$ with an infinity of terms.)

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The equation

$$e^x=x^y$$ is simply solved by

$$y=\frac{\ln x}x$$ but this is of no use.