# Number of solutions of the equation $a^{a^x}=x$ [duplicate]

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.

Thanks.

• See the post Solutions of $a^{a^x}=x$ for fixed $a \geq 0$ Dec 27, 2022 at 11:40
• 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.
– mcd
Dec 27, 2022 at 12:03
• @mcd I agree but :-/ Dec 27, 2022 at 12:25

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.