# Intuition for why equations of the form $k^x=x^c$ are not solvable trivially?

It recently occurred to me that I did not know how to solve equations of the form $k^x=x^c$ for any two constants $k$ and $c$. After much pain in algebraically manipulating the equation (using logarithms) I confirmed the x is trapped in an exponent and cannot be extracted without some external knowledge.

Finding the roots of the derivative creates a similar problem.

I then did some research and discovered Lambert's W function, which I haven't the mathematical maturity to understand deeply. (Solve $2^x=x^2$).

I'm looking for some rigorous intuition towards why without numerical methods or Lambert's W function the equation is not solvable for x, or more specifically, a proof as to why this class of function is unsolvable using only logs and rationals? The best answer I have for myself is that because x is both a power and an exponent, either extracting x with a log or removing the exponent with a rational will keep the other side of the equation unsimplified. What is the mathematical language with which to express this idea?

As an aside, I'm also wondering what area of mathematics considers issues such as these (this one being on the simpler side).

• I'm really sure how one would say this, but when $x$ appears in different 'levels' of exponents, it is not algebraically solvable, and when it appears in levels that are not directly next to each other, then solving non-trivially with the Lambert W function is also not possible. – Simply Beautiful Art Sep 22 '16 at 15:09
• These kind of equations are called transcendental equations, you might want to google on that word. – Nigel Overmars Sep 22 '16 at 15:15
• Why would they be solvable "trivially" whatever that means? Something being "trivial" is the exception, not the norm. You might as well ask for "intuition" for why math is hard. – Najib Idrissi Sep 22 '16 at 15:17
• "rigorous intuition"... I love this! – Alex M. Sep 23 '16 at 17:29

In general, functions are terrible, horrible, no good, very bad things, and you should never expect to be able to solve an equation like $f(x)=g(x)$. The essential reason for this is because there are way more real numbers than you think... so many, in fact, that if a real number is picked at random, it is an almost-certainty that you would not even be able to write that number down. So all I have to do is cook up a couple of functions which intersect at that point, and you're sunk.
Note that $k^x=x^c$ can be put into $e^{x\log k}=e^{c\log x}$ and so it is the same as $x\log k=c\log x$ and setting $y=\frac{1}{x}$ one find $$\frac{\log k}{c}=-y\log y.$$ Then setting $y=e^s$ one obtain a more familiar form from which one can use the definition of Lambert W function$^\dagger$, in fact $$-\frac{\log k}{c}=se^s$$ and using the definition of $W$ you have $s=W(-\frac{\log k}{c})$ and so $y=e^W(-\frac{\log k}{c})$ and finally $$x=e^{-W(\frac{-\log k}{c})}.$$
$^\dagger$ The Lambert W function is defined as the inverse of $f(x)=xe^x$ meaning that $$x=f(W(x))=W(f(x))=W(xe^x)=W(x)e^{W(x)}.$$