Solve the equation $x(\log \log k - \log x) = \log k$ I want to solve this equation by expressing $x$ in function of $k$. Is it possible?
Thanks.
 A: Yes, but you're gonna need the Lambert W function.
\begin{align}
x(\log \log k - \log x) &= \log k\\
\log x - \log \log k &= \frac{-\log k}x\\
\frac x{\log k} &= e^{-(\log k)/x}\\
-1 &= \frac{-\log k}{x}\cdot e^{-(\log k)/x}\\
W(-1) &= \frac{-\log k}{x}\\
x &= \frac{-\log k}{W(-1)}\\
\end{align}
Note that $W(-1)$ is a complex number.
A: there are no real solutions. Otherwise let $\log k=a$. Then we have $x(\log a-\log x)=a$ and so $\log(a/x)=a/x$ or $a/x=e^{a/x}$. No real solutions.
A: The equation is easily rewritten as $$\log\left(\frac{\log k}x\right) = \frac{\log k}x,$$
or $\log(z)=z$ or $e^z=z$.
As $\log(z)<z$, there are no real solutions, so no, it is not possible.

As said by others, there's a complex solution $z=W(-1)$ found using Lambert's function.
Alternatively, set $z=a+ib$. Then
$$e^a\cos(b)=a,\\e^a\sin(b)=b.$$
and, eliminating $a$,
$$a=b\cot(b)=\log(b\csc(b)).$$
This transcendental equation has an infinity of solutions.
In any case,
$$x=\frac{\log(k)}z.$$
A: Use $y=\frac x{\ln k}$ and you get:
$$
x(\ln(\frac x {\ln k})) = -\ln k\\
(\ln k) y \ln y= -\ln k\\
y\ln y =-1\\
ye^y=1/e
$$
The last equation can be solved like shown here:

$$
    x^x=z\,\\
    \Rightarrow x\ln x = \ln z\,\\
    \Rightarrow e^{\ln x} \cdot \ln x = \ln z\,\\
    \Rightarrow \ln x = W(\ln z)\,\\
$$
  or, equivalently,
  $$
    x=\frac{\ln z}{W(\ln z)}, 
$$

so  $\displaystyle x=\frac{\ln e^{-1}\ln k }{W(\ln e^{-1})}=\frac{-\ln k}{W(-1)}$
A: differentiating both sides with respect to x and taking  $$loglogk = p$$
$$d(px)/dx - d(xlogx)/dx = 0$$
$$p - x*d(logx)/dx - logx = 0$$
$$p - 1 - logx = 0$$
$$logx = p - 1$$
$$x = e^p/e$$
$$x = (logk)/e$$
