How to extract $x$ from the exponential equation $(x-1)\cdot\left(2^{1/x}-1\right)=k$ when $k\in(0,1)$ and $x\gt 1$?

Question

I spent quite a time searching the internet to find a way to extract $x$ from the exponential equation $(x-1)\cdot\left(2^{1/x}-1\right)=k$ when $k\in(0,1)$ and $x\gt 1$

I would appreciate any hints.

Answer 1

As @Tyma Gaidash commented, except using a generalized Lambert function , rewriting $$e^{-\log(2)\,t} =\frac{(1-k) t+1}{t+1} \qquad \text{where} \qquad t=-\frac 1x$$ (look at equation $(4)$ in the linked paper), you cannot obtain an explicit solution for the zero of function $$f(x)= (x-1)\left(2^{\frac{1}{x}}-1\right)-k$$ and numerical methods will be required.

However, you can have decent approximations using a simple $[2,2]$ Padé approximant built around $x=1$.

$$(x-1)\left(2^{\frac{1}{x}}-1\right) \sim \frac {(x-1) + a (x-1)^2}{1+b (x-1)+c(x-2)^2} \tag 1$$ $$a=\frac{-6-13 \log ^2(2)+18 \log (2)}{3 (3 \log (2)-2)}\quad b=\frac{\log ^3(2)}{3 (3 \log (2)-2)}\quad c=\frac{-6+5 \log ^2(2)+6 \log (2)}{3 (3 \log (2)-2)}$$

This means that you are left with a quadratic equation $$(a-c k)\,(x-1)^2 + (1-b k)\,(x-1)-k=0$$

Trying with this first approximation

$$\left( \begin{array}{ccc} k & \text{estimate}& \text{solution} \\ 0.05 & 1.05373 & 1.05373 \\ 0.10 & 1.11617 & 1.11617 \\ 0.15 & 1.18970 & 1.18970 \\ 0.20 & 1.27765 & 1.27765 \\ 0.25 & 1.38487 & 1.38487 \\ 0.30 & 1.51866 & 1.51866 \\ 0.35 & 1.69058 & 1.69056 \\ 0.40 & 1.92004 & 1.91999 \\ 0.45 & 2.24243 & 2.24231 \\ 0.50 & 2.72982 & 2.72946 \\ 0.55 & 3.55501 & 3.55385 \\ 0.60 & 5.26268 & 5.25803 \\ 0.65 & 10.9352 & 10.8990 \end{array} \right)$$

If you want more accurate solution, use Newton method with the estimate and apply $$x_{n+1}=x_n- \frac{f(x_n)}{f'(x_n)}$$ where $$f'(x)=-\frac{2^{\frac{1}{x}} (x-1) \log (2)}{x^2}+2^{\frac{1}{x}}-1$$

Edit

The model given by $(1)$ (remember that is is developed only on the basis of properties around $x=1$) is good for $0 \leq k \leq 0.4$. It can be improved doing the same kind of work around $\infty$

$$(x-1)\left(2^{\frac{1}{x}}-1\right) \sim \log(2) \frac {1+ \frac a x + \frac b {x^2}}{1+ \frac c x + \frac d {x^2}} \tag 2$$

$$a=\frac{-120+\log ^3(2)-12 \log ^2(2)+60 \log (2)}{10 \left(12+\log ^2(2)-6 \log (2)\right)}\qquad \qquad b=\frac{\log ^2(2)}{60}$$ $$c=-\frac{2 \log (2) \left(15+\log ^2(2)-7 \log (2)\right)}{5 \left(12+\log ^2(2)-6 \log (2)\right)}\qquad \qquad d=\frac{\log ^2(2)(\log(2)^2-8 \log (2)+20) }{20 \left(12+\log ^2(2)-6 \log (2)\right)}$$ and again a quadratic equation in $\frac 1x$.

$$\left( \begin{array}{ccc} k & \text{estimate}& \text{solution} \\ 0.69 & 144.321 & 144.321 \\ 0.68 & 34.8560 & 34.8560 \\ 0.67 & 19.9716 & 19.9716 \\ 0.66 & 14.0672 & 14.0672 \\ 0.65 & 10.8990 & 10.8990 \\ 0.64 & 8.92261 & 8.92261 \\ 0.63 & 7.57178 & 7.57178 \\ 0.62 & 6.58997 & 6.58997 \\ 0.61 & 5.84404 & 5.84404 \\ 0.60 & 5.25802 & 5.25803 \\ 0.59 & 4.78542 & 4.78543 \\ 0.58 & 4.39617 & 4.39617 \\ 0.57 & 4.06996 & 4.06997 \\ 0.56 & 3.79259 & 3.79260 \\ 0.55 & 3.55384 & 3.55385 \\ 0.54 & 3.34614 & 3.34615 \\ 0.53 & 3.16378 & 3.16380 \\ 0.52 & 3.00238 & 3.00240 \\ 0.51 & 2.85851 & 2.85853 \\ 0.50 & 2.72943 & 2.72946 \\ 0.49 & 2.61298 & 2.61301 \\ 0.48 & 2.50738 & 2.50741 \\ 0.47 & 2.41116 & 2.41119 \\ 0.46 & 2.32313 & 2.32316 \\ 0.45 & 2.24227 & 2.24231 \\ 0.44 & 2.16773 & 2.16778 \\ 0.43 & 2.09881 & 2.09886 \\ 0.42 & 2.03487 & 2.03492 \\ 0.41 & 1.97539 & 1.97546 \\ 0.40 & 1.91993 & 1.91999 \end{array} \right)$$

Answer 2

a) Impossibility of algebraic solving

Because, in the general case, the equation is a polynomial equation in dependence of algebraically independent monomials ($x,2^\frac{1}{x})$, the equation cannot be solved for $x$ by rearranging it by applying only finite numbers of elementary functions/operations we can read from the equation.

Other tricks, Special functions, numerical or series solutions could help.

b) Lambert W

Your equation is an exponential-polynomial equation.

$$(x-1)\left(2^\frac{1}{x}-1\right)=k$$ $$(x-1)\left(e^{\ln(2)t}-1\right)=k$$ $x\to\frac{\ln(2)}{t}$: $$\left(\frac{\ln(2)}{t}-1\right)\left(e^t-1\right)=k$$ $$\left(\frac{\ln(2)}{t}-1\right)e^t-\frac{\ln(2)}{t}+1=k$$ $$e^t=\frac{(k-1)t+\ln(2)}{-t+\ln(2)}$$ $$\frac{-t+\ln(2)}{(k-1)t+\ln(2)}e^t=1$$ $$\frac{t-\ln(2)}{(k-1)t+\ln(2)}e^t=-1$$ $$\frac{t-\ln(2)}{t+\frac{\ln(2)}{k-1}}e^t=-(k-1)$$

We see, your equation cannot be solved in terms of Lambert W in the general case. But it can be solved in terms of Generalized Lambert W:

$$t=W\left(^{\ \ \ln(2)}_{-\frac{\ln(2)}{k-1}};-(k-1)\right)=-W\left(_{-\ln(2)}^{\ \ \frac{\ln(2)}{k-1}};-\frac{1}{k-1}\right)$$

$$x=\ln(2)\ W\left(^{\ \ \ln(2)}_{-\frac{\ln(2)}{k-1}};-(k-1)\right)^{-1}=-\ln(2)\ W\left(_{-\ln(2)}^{\ \ \frac{\ln(2)}{k-1}};-\frac{1}{k-1}\right)$$

$-$ see the references below.

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

[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: 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