I was thinking how to solve:

  • If $x^{(x-1)^2}=2x+1$, find $x-\frac{1}{x}$
  • Solve $x^{x-x^2+13} = x^2-12$

I noticed that in both problems, the linear part can be constructed in the quadratic exponent, I tried a few change of variable and nothing.

  • $\begingroup$ ok, for first poblem $x=1+\sqrt{2}$. For the second $x=3, x=4$ by inspection. $\endgroup$ – Luis Felipe Mar 30 '20 at 15:13
  • $\begingroup$ I think you mean $-3$, not $3$. $\endgroup$ – See Hai Mar 30 '20 at 15:26
  • $\begingroup$ @ONGSEEHAIHCI yeah, missed the minus sign haha $\endgroup$ – Luis Felipe Mar 30 '20 at 15:28
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    $\begingroup$ @user134338 what do you think of my solution? $\endgroup$ – See Hai Mar 30 '20 at 15:28
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    $\begingroup$ thank you! But here's an observation for $1$: The solutions are actually equivalent to solving the equation: $x^2-2x-1=0$. $\endgroup$ – See Hai Mar 30 '20 at 15:38

Here's a possible approach to the second problem:

Observe that $x-x^2+13=-(x^2-12)+x+1$, so let $a=x^2-12$.

Now, we wish to solve the equation $x^{-a+x+1}=a$. Note that $x=0$ is not a solution; this is by inspection.

It is easy to see that $x=a$ actually satisfies the preceding equation. Now, we argue that all other values of $x$ does not work:

$x>a \Rightarrow x-a >0 \Rightarrow x-a+1>1 \Rightarrow x^{x-a+1}>a^{x-a+1} >a \ \forall \ x$, if $x \in \mathbb{R^+}$. So, equality does not hold.

Likewise, $x<a \Rightarrow x-a <0 \Rightarrow x-a+1<1 \Rightarrow x^{x-a+1}<a^{x-a+1} <a \ \forall \ x$, if $x \in \mathbb{R^+}$, and $x-a+1 \geq 0$. Otherwise, if $x-a+1<0$, $ x^{x-a+1}>a^{x-a+1} >a$. So equality never happens.

Thus, given that $x=a$, we have $x^2-12=x \Rightarrow (x+4)(x+3)=0 \Rightarrow x=4$ or $ x=-3$, and these are the two solutions to our equation, so we are done.

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    $\begingroup$ it is very clever ! $\endgroup$ – Luis Felipe Mar 30 '20 at 15:37
  • $\begingroup$ @ONG SEE HAI HCI $-3$ is not root, by the domain. The domain is $x>0$. $\endgroup$ – Michael Rozenberg Mar 30 '20 at 16:18
  • $\begingroup$ sorry, can you elaborate? $\endgroup$ – See Hai Mar 30 '20 at 16:19

Following the method of ONG SEE HAI HCI , we have $A = 2x+1$ then:

$$x^{(x-1)^2}=x^{x^2+2-(2x+1)}=2x+1$$ then $x^{x^2+2-A} = A $ with only solution $x=\sqrt{A}$.

This is because $x^{x^2+2-A}$ is increasing.

Then $x=\sqrt{A} \rightarrow x^2 = 2x+1 \rightarrow x = 2+ \frac{1}{x} \rightarrow x-\frac{1}{x} = 2$

  • $\begingroup$ I am sorry, is it obvious that it's increasing? $\endgroup$ – See Hai Mar 30 '20 at 16:22
  • $\begingroup$ this is because its equal to $x^{(x-1)^2}$ $\endgroup$ – Luis Felipe Mar 30 '20 at 16:27
  • $\begingroup$ @ONGSEEHAIHCI same argument can be used for your answer $\endgroup$ – Luis Felipe Mar 30 '20 at 16:27
  • $\begingroup$ I see your point now. $\endgroup$ – See Hai Mar 30 '20 at 16:28

A hint to the first problem.

Prove that for $x\geq2$ the function $f(x)=(x-1)^2\ln{x}-\ln(2x+1)$ increases.

Indeed, for $x\geq2$ we obtain: $$f'(x)=2(x-1)\ln{x}+\frac{(x-1)^2}{x}-\frac{2}{2x+1}>0.$$

Thus, since $f(2)<0$ and $\lim\limits_{x\rightarrow+\infty}f(x)=+\infty,$

we see that our equation has an unique root on $[2,+\infty).$

Now, easy to check that $1+\sqrt2$ is a root.

Indeed, $$(1+\sqrt2)^{(1+\sqrt2-1)^2}-2(1+\sqrt2)-1=(1+\sqrt2)^2-3-2\sqrt2=0.$$

The second equation we can solve by the similar way. I got that $4$ is a root.

  • $\begingroup$ I got uniqueless of root between 2 and 3 byother argumetns. But how you got $1+\sqrt{2}$ algebraically? $\endgroup$ – Luis Felipe Mar 30 '20 at 14:40
  • $\begingroup$ @user134338 I added something. See now. $\endgroup$ – Michael Rozenberg Mar 30 '20 at 14:43
  • $\begingroup$ that is verification, but how you known that $1+\sqrt{2}$ will be useful? $\endgroup$ – Luis Felipe Mar 30 '20 at 14:46
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    $\begingroup$ You proved uniqueless of solution, and then the value $1+\sqrt{2}$ appears magically solving the equation. How does you think on $1+\sqrt{2}$ ? $\endgroup$ – Luis Felipe Mar 30 '20 at 14:52
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    $\begingroup$ But, as what @user134338 mentioned earlier, how did you even arrive at 1+$\sqrt(2)$ in the first place? It seems like a mysterious number that popped out of nowhere $\endgroup$ – See Hai Mar 30 '20 at 14:53

I think I understand what @Michael Rozenberg is trying to say now. Basically, he views the equation $f(x)=g(x)$ as the point(s) of intersection between the two functions $y=g(x)$ and $y=f(x)$. For example, for the second problem, to solve $x^{x-x^2+13}=x^2-12$, he says that we are essentially finding the points of intersection of $2$ functions, $f(x)=x^{x-x^2+13}$ and $g(x)=x^2-12$. But under such an interpretation, we have to take the domains of the functions into consideration, which means that there might be solutions to the original solution that were being omitted.

So here's how to proceed: First note that $D_f=x \in \mathbb{R^+_{0}}$, and $D_g=\mathbb{R}.$ Now, $x^2-12=-(x-x^2+13)+x+1$, so let $x-x^2+13=A$. Thus, the original equation reduces to: $x^A=-A+x+1$, $x \geq 0$. Then, treating $A$ as a constant, we have the solution to this new equation to be $A=1$; this can be seen by inspection. It is easy to prove that no other values of $A$ will give us equality:

Let $h(x)=x^A+A-x-1,x\geq0$. When $x=A$, we have $h(x)=0$. Also, $h'(x)=Ax^{A-1}-1$. If $A>1$, $h'(x)=Ax^{A-1}-1 > x^{A-1}>x^0-1>0,$ where the second last inequality follows since $x$ is non-negative. Thus, $h(x)$ is strictly increasing and $\geq 0$ $\forall \ x \in \mathbb{R^+_{0}} \Rightarrow \ f(x)>g(x) \ \forall \ x \in \mathbb{R^+_{0}}$. Likewise, if $A<1, h'(x)=Ax^{A-1}-1<x^{A-1}-1<x^0-1<0$ implies $h(x)$ is strictly decreasing and $ \leq 0$$\forall \ x \in \mathbb{R^+_{0}} \Rightarrow \ f(x)<g(x) \ \forall \ x \in \mathbb{R^+_{0}}$.

Thus, given $A=1$, let $x-x^2+13=1 \Rightarrow x=-3 $ or $x=4$, but we reject the former solution since we have to take into account the domain of $f(x)$, $D_f$.


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