When I studied math at my school (I was 16) I was studying system of linear equations, and equations of first or second order and I liked it very much (and I was good to xD) but one day I asked to myself how to solve equation of this kind $ax=a^x$ and …I had no idea, I have felt completely disorientated.
Now math is only an hobby for me (even if I don't study it at school I love it) so I tried again to solve this problem, using "new concept" I have "learnt" since I left school but still can't find a solution even using logarithm (aka even using a calculator).
I noticed that $a^x-ax=0$ so I just need to find the zeros of the functions $f_a(x)=a^x-ax$ but I don't know how to solve this.
My second attempt was to using a graphical method plotting $ax=y$ and $a^x=y$ and finding the intersection, but this doesn't satisfy me.
My last effort was to "unify" the two $x$'s and put them on the same side of the equation:
$$a^x=ax$$
$$\log_a(a^x)=\log_a(ax)$$
$$x=1+\log_a(x)$$
$$a^{x-1}=x$$
$$a=x^{1/(x-1)}$$
here I'm lost and I am starting to doubt that is possible to "bring" the $x$s using the "standard" algebra rules, to one side of the equation and get one $x$ "alone".
Maybe I need more powerful theoretic tools to face this problem, but I don't understand why I can't find a way to "reduce" it to a simple form with easy operations $(+,\cdot)$ and powers.
a) Maybe that means that the solution isn't even irrational? Why at this point all become harder?
b) How can I face this problem? How to find the solution? Which are the extra concepts I need? I accept complete answers (even if I wanted only hints because I love to solve problems on my own, but I think this is above my level)
I'm sorry for my terrible English but I'm using a translator. Thanks in advance.
Best Answer
We need to bring out a non-elementary (termed "special") function for this kind of problem; the one intended for this type of situation is the Lambert W function. It is defined to be the inverse function of $xe^x$ - that is, it satisfies the functional equation $W(x)e^{W(x)}=x$.
Now we wish to solve $ax=a^x$. Fix $a$, so that we're solving for $x$ in terms of $a$. Rearrange:
$$ax=e^{(\ln a)x}\iff xe^{-(\ln a)x}=a^{-1}\iff \color{Blue}{-(\ln a)x} e^{\color{Blue}{-(\ln a)x}}=\color{DarkGreen}{-(\ln a)/a}$$
$$\iff \color{Blue}{-(\ln a)x}=W\left(\color{DarkGreen}{-\frac{\ln a}{a}}\right)\iff x=-\frac{1}{\ln a}W\left(-\frac{\ln a}{a}\right).$$
Inverse functions can trip a lot of people up when they're new to the concept, and in particular, those that are inverse functions which are non-elementary (not able to be expressed with a finite number of the basic four operations combined with roots and powers and exponentials) can be pretty trick and take a bit of getting used to. The Lambert W function falls into this category occasionally, which is why I provide a full solution as an introduction on how to manipulate it.