Solve the equation $x(\log \log k - \log x) = \log k$

Question

I want to solve this equation by expressing $x$ in function of $k$. Is it possible?

Thanks.

Answer 1

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.

Answer 2

differentiating both sides with respect to x and taking
$$loglogk = p$$ $$d(px)/dx - d(xlogx)/dx = 0$$ $$p - x*d(logx)/dx - logx = 0$$ $$p - 1 - logx = 0$$ $$logx = p - 1$$ $$x = e^p/e$$ $$x = (logk)/e$$

Answer 3

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.

Answer 4

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 $\displaystyle x=\frac{\ln e^{-1}\ln k }{W(\ln e^{-1})}=\frac{-\ln k}{W(-1)}$

Answer 5

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.

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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,

$$x=\frac{\log(k)}z.$$