exact solution to $x+x^x=3$

Question

I'm trying to find the exact solution to the equation $x+x^x=3$. I know the answer is approximately 1.4, but what is its formal definition? I understand it may not be from the result of a real function, since the exact solution to its brother, $x^x=3$, requires use of the Lambert W function for its formal definition ($x=e^{W(\ln 3)} $), which cannot be expressed in elementary terms. I have tried looking this equation up on Wolfram alpha but it can only approximate $x$. What would be the exact definition of this variable?

Answer 1

Irrational numbers are usually defined as solutions of equations, for example $x^2=2$, as limits, say $\sum_{k=0}^\infty(-1)^k\frac{4}{2k+1}$ or $\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{1/n}$, or by introducing new notation, say $\sqrt{2},\pi$ or $e$. You stated the definition as a solution of an equation. Since introducing new notation is not helpful in this case, I only propose the following two alternatives. Let $x^*\in(0,\infty)$ be the solution of $x+x^x=3$.

1. Limit: Consider the sequence given by $a_0=1$ and $a_{k+1}=\frac{3}{1+a_k^{a_k-1}}$ for $k\ge 0$. The sequence $(a_k)_k$ converges to the solution of the equation, i.e. $x^*=\lim_{k\rightarrow\infty}a_k$.
2. Using the inverse function: Let $f:[1,\infty)\rightarrow(0,\infty)$, $x\mapsto x+x^x$, and notice that $f$ is strictly increasing since $f'(x)=1+(\ln(x)+1)x^x>2$, so the solution of the equation is $x^*=f^{-1}(3)$.

These would be three common ways (equation, limit, function value) to formally define the number $x^*$. I hope this helps!

Answer 2

$$x+x^x=3$$

We see, this equation is a polynomial equation of more than one algebraically independent monomials ($x,x^x$) and with no univariate factor. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (elementary operations) we can read from the equation.

$$x+e^{x\ln(x)}=3$$ $$e^{x\ln(x)}=3-x$$ $$1=(3-x)e^{-x\ln(x)}$$ $$(3-x)e^{-x\ln(x)}=1$$

We see, we cannot solve the equation in terms of Lambert W because of the exponent of the exponential.

$x\to 3-t$: $$te^{(t-3)\ln(3-t)}=1$$

We see, the equation can be solved in terms of Hyper Lambert W.

$$G((t-3)\ln(3-t)e^{-t};t)=1$$ $$t=HW((t-3)\ln(3-t)e^{-t};1)$$ $$x=3-HW((t-3)\ln(3-t)e^{-t};1)$$

So we have a closed form for $x$, and the representations of Hyper Lambert W give some hints for calculating $x$.

Galidakis, I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997

Galidakis, I. N.: On some applications of the generalized hyper-Lambert functions. Complex Variables and Elliptic Equations 52 (2007) (12) 1101-1119