Hints on solving $x^{2x}-(x^2+x)x^x+x^3=0$

Question

Solve this equation over $\mathbb{R}^+$: $x^{2x}-(x^2+x)x^x+x^3=0$

I’ve been trying to solve this exponential equation but can’t get the answer because normal substitution ($y=x^x$) isn’t working. Any tips/hints that don’t use logs? (the section of the book I got this from is before the introduction of logarithms). Thanks.

Answer 1

What you have is a transcendental equation in $x$, meaning that it is not a polynomial in $x$ (i.e., generated through arithmetic operations $+,\times,-,\div$ only) nor even an algebraic equation (i.e., corresponding to a function that would in turn solve another polynomial). Compare this with equations such as $x-\cos x=0$ or $x\cdot e^x=1$ (e.g., those mixing different classes of elementary functions).

But beware: transcendental functions don't tend to admit general closed-form solutions or inverses (written in terms of the functions we like to use), and I wouldn't suggest approaching them with the mindset that they should have nice solutions. What you can often hope for, however, is (1) to be lucky enough to already have, from the offset, something with easy solutions (e.g., $x=0$ or $x=1$), (2) to characterize the intended solution by proving its properties (e.g., existence, bounds, rationality,...), (3) to approximate the solution, if it does exist, or (4) to define a new function that describes the operation of the inverse (as in the case of the Lambert W). I find that novice mathematicians are often drawn to option (4) since it seems to reveal a powerful hack for undoing tricky equations, but I think it's actually less useful than it looks. You're effectively just assigning a symbol to "the solution", which is little more insightful than writing $x=f^{-1}(y)$.

Here, we cannot easily invert the expression (whether we factorize it or not). We are lucky enough that it has $x=1$ as a closed-form solution, but if it had been the similar equation $x^{x}+x-3$ with root $x=1.399\ldots$, it wouldn't (all we'd be able to do would be to define a new inverse function or approximate it). Then, once we have identified our solutions, closed-form or not, we can prove that there are no others (i.e., uniqueness) by observing where $f(x)$ is increasing/decreasing.

Answer 2

There are only three obvious solutions $x=-1,x=1$ and $x=2$

The other intervals are inconsistent with the sign of the simplified equation:

$x^{2x}-(x^2+x)x^x+x^3$ = 0 { expand } $x^3+x^{2x}-x^{1+x}-x^{2+x}=0$ {a} $1+x^{-3+2x}-x^{-2+x}-x^{-1+x}=0$ Divide {a} by $x^3$

Simplified to $-x^{x-2}-x^{x-1}+x^{x^2-3}=-1$ (1)

for example if x>2

$-x^{x-2}-x^{x-1}+x^{x^2-3}$ is always positive sign where is inconsistent with the equation simplified (1):

Cause: when x>2

$-x^{x-2}-x^{x-1} < x^{x^2-3}$ So $-x^{x-2}-x^{x-1}+x^{x^2-3}$ is always positive.

So No solution for x>2.

You can continued the demonstration with the intervals : 1<x<2 , 0<x<1, x=0 , -1<x<0 , x<-1 . All this intervals gives no solutions in Reals.

...