2

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.

Angelo
  • 12,328
Nickferna
  • 31
  • Sorry for the bad image, yes you got it right. I could guess 0 and 1 but am not sure how to proceed in a general manner. – Nickferna Feb 18 '23 at 22:18
  • 15
    An alternate form is $(x^x-x)(x^x-x^2)=0$ – Тyma Gaidash Feb 18 '23 at 22:20
  • 1
    Notice that $$t^2 - (a + b) t + a b = (t - a) (t - b)$$ take $t = x^x$, $a = x^2$ and $b = x$. – Essaidi Feb 18 '23 at 22:53
  • Thanks for the answers. How can I learn identities like the one one Essaldi cited? – Nickferna Feb 18 '23 at 23:03
  • The identity is just from factoring trinomials. – Тyma Gaidash Feb 18 '23 at 23:04
  • 1
    The only real solution of $x^x-x=0$ is $x=1$ and the only real solutions of $x^x-x^2=0$ are $x=1$ and $x=2$. Hence all the real solutions of the OP’s equations are $x=1$ and $x=2$. – Angelo Feb 18 '23 at 23:14
  • @Angelo, what you write is correct, but I wonder whether it's possible to prove that those are the only real solutions, using only the parts of math that OP knows about. It's not clear to me whether OP has done any Calculus, for example. – Gerry Myerson Feb 19 '23 at 00:25
  • For real $x$, we have $x^{2x}=\left(x^x\right)^2$. Therefore your equation is a polynomial equation of the algebraically independent monomials $x^x$ and $x$. Therefore we don't know how we can invert the elementary function on the left-hand side of the equation only by elementary functions. But we can solve for $x^x$: $x^x=x$ and $x^x=x^2$. Therefore $x_{1,2,3}=-1,1,2$ are solutions. – IV_ Feb 19 '23 at 13:10

2 Answers2

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.

Jam
  • 10,325
  • 2
    The whole point of the question (or so one must assume) is that the term can be factored, leading to a very simple solution indeed. – David G. Stork Feb 18 '23 at 23:31
  • @DavidG.Stork Meh. Once you've factored it, you're not really solving each factor algebraically, you're recognizing their solutions by inspection (which in principle, you could've done to begin with). Hence my point about similar expressions (if we were really solving $x^x-x$ algebraically, we should be able to solve $x^{x}-x-1$ algebraically too). The OP's remarks in the question body about substitutions and logarithms made it apparent that they were looking for an algebraic way of solving it, which I thought it wise to suggest against. – Jam Feb 18 '23 at 23:38
0

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.

...

Reda.Kebbaj
  • 101
  • 4
  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Feb 19 '23 at 22:04
  • 1
    Here is advice on formatting mathematics on this website: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference but I don't see how your (first) "simplified equation" is equivalent to the equation(s) OP wants to solve. – Gerry Myerson Feb 19 '23 at 23:50
  • $Gerry Myerson Thanks for the link "formatting". I modified my answer to explain how "simplified equation" is equivalent to the OP want to solve. – Reda.Kebbaj Feb 20 '23 at 14:26