Find all x such that : $x^{x^{x^3({x^{x^{3}}+1)}+3}}=3^{81}$

Question

This was the question :

Find all x such that : $x^{x^{x^3({x^{x^{3}}+1)}+3}}=3^{81}$

By observation (error and trial) I was able to find that $x=\sqrt[3]3$ is a solution , also , I was able to prove that there are no other solutions for positive x . As , for $x\in (0,1)$ there can not be any solution as LHS is less than $1$ and RHS is greater than $1$ , and for $x\in (1,\infty)$ the function is strictly increasing so there can be only one solution in that range .

But I am not able to say anything certainly about negative x , if I plot this funtion on desmos , it does not plot anything in the second quadrant ( Clearly the function is not defined for almost all negative x , but there are some negative values of x for which the function is defined , I didn't understand why they were also not plotted at least , I saw the same thing with the function $x^x$ , is it that these functions are just not defined for negative input ? ).

So , if anyone could help me in finding the negative solution (if there is any ) , or teach me any way to properly prove (I mean it is very intuitive that there is no negative solution but I couldn't support it with a solid reasoning) that there are no solutions for negative x , that would really be a great help !

Thanks!

Answer 1

Well, let $y=x^{x^3}$, we have $$x^{x^{x^3(x^{x^3}+1)+3}}=x^{x^{x^3(x^{x^3}+1)}\cdot x^3}=(x^{x^3})^{(x^{x^3})^{(x^{x^3}+1)}}=y^{y^{y+1}}$$ let $z=y^y$, we have $$y^{y^{y+1}}=y^{y^y\cdot y}=(y^y)^{y^y}=z^z=3^{81}=(3^3)^{27}=27^{27}$$ So, we can get $z=27$ and thus $y=3$ and thus $$x^{x^3}=3 \implies x^{3x^3}=3^3 \iff (x^3)^{x^3}=3^3$$ and we can let $m=x^3$ and get $m^m=3^3 \implies m=3 \implies x=\sqrt[3]{3}$

Note: I did say several times that $a^a=3^3\implies a=3$ or $b^b=27^{27} \implies b=27$, and yes it's true because if $$a^a=b^b \implies a\ln(a)=b\ln(b)$$ and because $f(x)=x \ln(x)$ is increasing for all $x \ge 1$ $$a>b \implies \ln(a)>\ln(b) \implies a\ln(a)>b\ln(b)$$ we get that $f$ is injective for all $x \ge 1$ and thus $a \ln(a) = b \ln(b) \implies a=b$ for all $a,b \in [1, \infty)$