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We're given 2 constants $a,b\geq 1$ and we need to calculate x such that: $x=a\log_2(bx)$.

I thought of doing the following:

$$x=\frac{ba\log_2(\frac{ba}{a}x)}{b}=\frac{ba\log_2(ba\log_2(\frac{ba}{a}x))}{b}=\frac{ba\log_2(ba\log_2(ba\log_2(...)))}{b}$$

From the second transition each time I replace $x$ with $a\log_2(bx)$ and I do it infinite times.

And then we need to find $t$ such that:

$$t=ba\log_2(t)$$ $$\frac{t}{\log_2(t)}=ba$$

And we would get that: $$x=\frac{t}{b}$$

So I have 3 question:

1) Was the first infinite log transition even "legal"?

2) If it is then how can I solve the equation for $t$? Is there any inverse function for $\frac{t}{\log_2(t)}$?

3) If the answer to 1 or 2 is no, then is there any other way to solve this?

Tomer Wolberg
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    Cannot be solved by elementary methods. The solution can be written with the Lambert W function. https://en.wikipedia.org/wiki/Lambert_W_function – GEdgar May 17 '20 at 20:55

1 Answers1

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  1. It probably can be made legal by saying something about fixed points of function $t \to ba\log_2 t$, but we don't need it. After you got idea about $x = \frac{t}{b}$, you can substitute it to the original equation to get $t = ab\log_2 t$.

  2. Value of $t$ can't be expressed in elementary functions, but can be in Lambert W function. Just transform your equation on $t$ to form $t = e^{ct}$ and substitute $c$ to formula there.

mihaild
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