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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.

Karim Kamil
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1 Answers1

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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.

mcd
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    Is this enough for you to solve the problem, or would you like me to elaborate? – mcd Dec 27 '22 at 11:40
  • 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. – Tirterra Dec 27 '22 at 11:45