Given a positive real number $a≠1$, it is asked to determine according to the values of $a$ the number of real solutions of the equation $a^{a^x}=x$.

My try :

First any possible solution $x$ must be positive.

Then the equation is equivalent to

$$\lambda e^{\lambda x} = \ln(x)$$

Where $\lambda= \ln(a)$

Any advice on how to proceed from here would be great.


  • 1
    $\begingroup$ See the post Solutions of $a^{a^x}=x$ for fixed $a \geq 0$ $\endgroup$ Dec 27, 2022 at 11:40
  • 2
    $\begingroup$ It is a shame this question was closed: it would be useful to provide a simple solution to the problem for real solutions, as asked for: the reference questions is a sophisticated analysis of the complex solutions, which may not suit the poser of this query. $\endgroup$
    – mcd
    Dec 27, 2022 at 12:03
  • $\begingroup$ @mcd I agree but :-/ $\endgroup$ Dec 27, 2022 at 12:25

1 Answer 1


If you think about the shape of the graphs $y= \ln x$ and $y= \lambda e^{\lambda x}$, and their intersections with the axes, you can see that they might meet zero times (for large positive $\lambda$), twice (for suitably small positive $\lambda$) and once for any negative $\lambda$. They meet once for $\lambda = 0$ also, so the only question is "what is the boundary between the cases of positive $\lambda$?". This will be when the curves touch, and you can find the value of $\lambda$ for this by realising that the curves have both the same $y$ value and the same gradient at this point.

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    $\begingroup$ Is this enough for you to solve the problem, or would you like me to elaborate? $\endgroup$
    – mcd
    Dec 27, 2022 at 11:40
  • $\begingroup$ I was confused for a bit because I thought that he was asking for complex solutions, it would be a nice problem to find the amount of complex roots, if anyone knows how to proceed for complex solutions let me know. $\endgroup$
    – Tirterra
    Dec 27, 2022 at 11:45

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