I have this equation: $$ a \cdot 2^a = b$$ and need to find solutions for $a$.
And I know it's complicated and has to do something with the $W$ function.
But I've come up with a way which does not work for some-reason... and I would like you to have a look and help me find the mistake!
We can write any number $a = 2^k$ where $k$ isn't necessarily in $\mathbb{N}$ or $\mathbb{Z}$
so we have: $2^k \cdot 2^{2^k} = b$ which gives: $2^{2^k +1} =b $ and thus: $2^k + 1 = \log_2{b}$
And we get that $k = \log_2(\log_2b -1)$ and thus $a = 2^k = 2^{\log_2(\log_2b -1)} = \log_2b -1$
However this does not work! Why? Here is an example:
$a \cdot 2^a = 24$ In my example $a = \log_2(24) -1 \approx 3.58$ but it is actually $3$ ...
Edit: I am so dumb.. it should be $2^{2^k + k}$ ... Is there a way to solve this kind of equation?
Thank you!
$2^k \cdot 2^{2^k} = 2^{k + 2^k}$, not $2^{1+2^k}$.
EDIT: The solutions of $a 2^a = b$ are $$a = \frac{W(b \ln(2))}{\ln(2)}$$ where $W$ is a branch of the Lambert W function. If you want real solutions: there is none if $b < -1/(e \ln(2))$. If $b = -1/(e \ln(2))$, the only real solution is $-1/\ln(2)$. If $-1/(e \ln(2)) < b < 0$, there are two real solutions, one with the "$-1$" branch and one with the principal branch of $W$. If $b \ge 0$, the only real solution is with the principal branch.