solve a tricky equation

Question

I created a problem and i am looking for help in solving this equation. I do not know if any method exists except numerical methods. I tried to transform this equation to solve with Lambert W function but unsuccessfully.

$$x^{x+3}=125$$

Answer 1

If you make the problem more general, you are looking for the zero of function $$f(x)=x^{x+a}-b$$ which does not look very nice.

Better would be to consider instead $$g(x)=(x+a)\log(x)-\log(b)$$ Expanding as series around $x=1$ $$(x+a)\log(x)=(1+a)(x-1)+\sum_{n=2}^\infty (-1)^n\, \frac{1-a(n-1) }{n(n-1)}\,(x-1)^n$$

Use power series reversion to obtain $$x=1+\sum_{n=1}^\infty \frac{P_n(a)}{n!\, (a+1)^{2n-1}} \log^n(b)$$ where the first polynomials are $$\left( \begin{array}{cc} n & P_n(a) \\ 1 & 1 \\ 2 & a-1 \\ 3 & a^2-7 a+4 \\ 4 & a^3-25 a^2+67 a-27 \\ 5 & a^4-71 a^3+531 a^2-821 a+256 \\ 6 & a^5-181 a^4+3046 a^3-11606 a^2+12281 a-3125 \\ 7 & a^6-435 a^5+14666 a^4-113366 a^3+273141 a^2-217015 a+46656 \\ \end{array} \right)$$

For your specific example $(a=3,b=125)$, as a function of the order, the results are oscillating (which is not very good) and convergence is quite slow $$\left( \begin{array}{cc} n & x_n \\ 1 & 2.20708 \\ 2 & 2.57134 \\ 3 & 2.42477 \\ 4 & 2.39160 \\ 5 & 2.45300 \\ 6 & 2.44078 \\ 7 & 2.41835 \\ 8 & 2.43447 \\ 9 & 2.43869 \\ 10 & 2.42791 \\ \end{array} \right)$$

Probably better would be to use the very first iteration of any Newton-like method of order $n$ using $$x_0=1+\frac{\log(b)}{a+1}$$ This will give fully explicit formulae.

For your example $$\left( \begin{array}{ccc} n & x_1^{(n)} & \text{method} \\ 2 & 2.43115 & \text{Newton} \\ 3 & 2.43245 & \text{Halley} \\ 4 & 2.43225 & \text{Householder} \\ 5 & 2.43226 & \text{no name} \\ 6 & 2.43226 & \text{no name} \\ \end{array} \right)$$

Answer 2

$$x^{x+3}=125$$ $$e^{\ln(x)(x+3)}=125$$ $$\ln(x)(x+3)=3\ln(5)$$ $$\ln(x)\ x+3\ln(x)=3\ln(5)$$

We see, this equation is a polynomial equation of more than one algebraically independent monomials ($\ln(x),x$) 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.

$x\to e^t$: $$te^t+3t=3\ln(5)$$ $$te^t=3\ln(5)-3t$$ $$\frac{t}{3\ln(5)-3t}e^t=1$$

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

$$\frac{t}{t-\ln(5)}e^t=-3$$ $$t=W\left(^{\ \pm 0}_{\ln(5)};-3\right)$$ $$x=e^{W\left(^{\ \pm 0}_{\ln(5)};-3\right)}$$

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