How can we find Lambert W solution to $\dfrac {x\ln x}{\ln x+1}=\dfrac{e}{2}$?

Question

Find all real solutions: $$\frac {x\ln x}{\ln x+1}=\frac{e}{2}$$

Cross multiplication gives $$2x\ln x=\ln (x^e)+e$$ I didn't see any useful thing here. I tried solving this equation in WA. The interesting thing is that Wolfram provides $2$ real approximate solutions.$$x\approx 0.282858...$$ $$x\approx 2.71828...$$

When I looked at one of these solutions, I easily saw that the second solution was the Euler Number and immediately tried it in the equation.

$$\frac {e\ln e}{\ln e+1}=\frac {e}{1+1}=\frac {e}{2}$$

Indeed $x=e$ is a correct solution. One of the solutions is correct, but reaching that solution doesn't actually give us an idea of ​​the equation. Also I have no idea about the first solution.

Can we find Lambert W solution to $\dfrac {x\ln x}{\ln x+1}=\dfrac{e}{2}$ ?

Answer 1

There already were methods mentioned to find $x=e$, but for the other root, apply Lagrange reversion: $$\frac {x\ln x}{\ln x+1}=\frac e2\mathop\iff^{x=\sqrt[w]e} e(w+1)=2\sqrt[w]e\implies x=\frac1e-\sum_{n=1}^\infty\frac{\left(\frac 2e\right)^n}{n!}\left.\frac{d^{n-1}}{dw^{n-1}}\frac{e^\frac{n+1}w}{w^2}\right|_{-1}$$

Using a Maclaurin series and factorial power $n^{(m)}$ for the derivatives:

$$\left.\frac{d^{n-1}}{dw^{n-1}}\frac{e^\frac{n+1}w}{w^2}\right|_{-1}=\sum_{m=0}^\infty\frac{(-n-1)^m(-m-2)^{(n-1)}}{m!}$$

This uses a confluent hypergeometric function. We rewrite it as a Laguerre polynomial sum and apply a Laguerre $\operatorname L_n^1(x)$ integral representation,

$$\bbox[4px,border: 2.7px double #5ADEFF]{\frac {x\ln x}{\ln x+1}=\frac e2\implies x=e,x=1-\sum_{n=1}^\infty\left(\frac2e\right)^{n-1}\,_1\text F_1(n;2;-n)=\frac1e+\frac1{2\pi}\int_0^{2\pi}e^{it-e^{it}-1}\ln(1-2(e^{it}+1)e^{-e^{it}-it-2})dt}$$

$x=0.28285\dots$ is shown here

Answer 2

Making the problem more general, onsider that you look for the zero's of function $$f(x)=\frac{x \log (x)}{\log (x)+1}-k$$

If $k>0$ there are two roots $$0 <x_1 < \frac 1 e \qquad \text{and} \qquad 1 <x_2< \infty$$

To approximate the first root, make a series expansion around $x= \frac 1 e $

$$\frac{x \log (x)}{\log (x)+1}=-\frac{1}{e^2 \left(x-\frac{1}{e}\right)}-\frac{1}{2 e}+\frac{7}{12} \left(x-\frac{1}{e}\right)+\frac{e}{24} \left(x-\frac{1}{e}\right)^2-\frac{11e^2}{720} \left(x-\frac{1}{e}\right)^3+\frac{11 e^3 }{1440} \left(x-\frac{1}{e}\right)^4+O\left(\left(x-\frac{1}{e }\right)^5\right)$$ and use a power series reversion to obtain, as an estimate, $$x=\frac{1}{e}-\frac{1}{e^2 k}+\frac{1}{2 e^3 k^2}+\frac{1}{3 e^4 k^3}-\frac{19}{24 e^5 k^4}+\frac{1}{5 e^6 k^5}+\frac{757}{720 e^7 k^6}+O\left(\frac{1}{k^7}\right)$$

If you make $k=\frac e 2$, converted to decimals, the results is $x=0.282909$ while the exact solution, given by Newton method, is $x=0.282858$.

Using twice more terms in the original series would lead to $x=0.28285863$ to be compared to the "exact" $x=0.28285804$.

Answer 3

In this answer, we cover the question :

" How can the solution $x=e$ be reached without guessing ? "

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With a little manipulation on the equation, you can observe that the original equation is equivalent to :

$$ \begin{align}&2x\ln x=e(\ln x+1)\\ \iff &2x^2\ln x={\color{#c00}{xe}}(\ln x+1)\\ \iff &x^2\ln x^2=(\ln x+1){\color{#c00}{e^{\ln x+1}}}\\ \iff &\ln x^2\cdot e^{\ln x^2}=(\ln x+1)e^{\ln x+1}\end{align} $$

which implies, one possible real root comes from $\ln x^2=\ln x+1$ or equivalently $2\ln x=\ln x+1$ . Thus, we obtain that :

$$\ln x=1\implies x_1=e$$

is a solution .

Note that, we basically reduced the equation to $xe^x=ye^y$ .

It seems reasonable to assume that, the second real root can not be expressed in terms of closed-form expressions .

Answer 4

$$\frac{x \log (x)}{\log (x)+1}=k$$ $$x=e^t \quad \implies\quad \frac{e^t t}{t+1}=k\quad \implies\quad e^{-t}=\frac 1 k\, \frac t{1+t}$$ The solution is given in terms of the generalized Lambert function (have a look at equation $(4)$ in the linked paper).

Edit

We can also consider the more general case of $$x^a\,\, \frac{P_m(\log(x)}{Q_n(\log(x)}=k$$ and have $$e^{-at}=\frac 1 k\, \frac {P_m(t)}{Q_n(t)}=\frac 1 k\,\frac{\prod_{i=1}^m (t-r_i) }{\prod_{i=1}^n (t-s_i) }$$

Answer 5

$$\frac{x\ln(x)}{\ln(x)+1}=\frac{e}{2}$$

$$x\ln(x)-\frac{1}{2}e\ln(x)-\frac{1}{2}e=0$$

We see, the equation can be rearranged to a polynomial equation of more than one algebraically independent monomials ($x,\ln(x)$) and with no univariate factor. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (operations) we can read from the equation.

$$\frac{x\ln(x)}{\ln(x)+1}=\frac{e}{2}$$

$x\to e^t$:

for $t,x\in\mathbb{R}$:

$$\frac{t}{t+1}e^t=\frac{e}{2}$$

We see, we cannot solve this equation in terms of Lambert W, but in terms of Generalized Lambert W.

$$\frac{t-0}{t-(-1)}e^t=\frac{e}{2}$$

$$t=W\left(^{\pm 0}_{-1};\frac{e}{2}\right)$$ $$x=e^{W\left(^{\pm 0}_{-1};\frac{e}{2}\right)}$$

So we have a closed form for $x$, and the series representations of Generalized Lambert W give some hints for calculating $x$.
$\ $

[Mező 2017] Mező, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mező/Baricz 2017] Mező, I.; Baricz, Á.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

Answer 6

As noticed, it seems not possible obtain the solution by Lambert function, indeed by $x=e^y$ we have

$$\frac {x\ln x}{\ln x+1}=\frac{e}{2} \iff \frac {ye^y}{y+1}=\frac{e}{2}\iff y \;e^y=\frac{y+1}2\;e $$

which seems not reducible in the form $w \,e^w = \text{constant}$.

Anyway from the latter expression, since both sides are increasing, it is clear that only one solution exists for $y>0$ that is $y=1$.