Equations with variable Exponents

Question

I am struggling to find a solution to $x^{x-5}=5$, although clearly from plotting the graph of $f(x)=x^{x-5}-5$ I can see that there are two real solutions, but I have no idea how to evaluate them, or any other equations in the form $ax^{x\pm b}\pm c=0$. Hopefully someone can help me out here...

Answer 1

Problems of the form $x^x=a$ can be solved for $x$ using the Lambert W function. There is an example on the wikipedia.

Problems of the form $x^{x+y}=a$ cannot be solved using the Lambert W function. If the Lambert W function cannot solve an exponential equation, don't expect any logarithms or anything nice if you are looking for a closed form solution.

And believe me, the Lambert W function isn't even accepted as an elementary function, so this problem is probably beyond any closed form.

Answer 2

As already said in comments and answers, there is no analytical solution to this equation and numerical methods are required.

Considering $$f(x)=x^{x-5}-5$$ the first derivative is given by $$f'(x)=x^{x-5}\left(1-\frac{5}{x}+\log (x) \right)$$ which cancels at $$x_*=\frac{5}{W(5 e)}\approx 2.57141$$ $W(z)$ being Lambert function.

Using a calculator, you should find that $f(x_*)\approx -4.89911$ and the second derivative test shows that this is a minimum. So, two roots exist for $f(x)=0$. Graphing the function, you should notice that there is one root between $0$ and $1$ and another between $5$ and $6$.

Using Newton method with $x_0=0.5$, the successive iterates would be $$x_1=0.580369$$ $$x_2=0.647597$$ $$x_3=0.681899$$ $$x_4=0.688288$$ $$x_5=0.688468$$ which is the solution for six significant figures.

Doing the same using $x_0=5.5$, the successive iterates would be $$x_1=6.13041$$ $$x_2=5.95213$$ $$x_3=5.90836$$ $$x_4=5.90622$$ $$x_5=5.90621$$ which is the solution for six significant figures.

Edit

The problem can be made simpler if, instead of $f(x)$ we consider its logarithmic transform $$g(x)=(x-5)\log(x)-\log(5)$$ Repeating the calculations, we should have $$x_1=0.655752$$ $$x_2=0.687499$$ $$x_3=0.688467$$ and, for the second root $$x_1=5.92161$$ $$x_2=5.90623$$ $$x_3=5.90621$$

Answer 3

Equations of this type can be solved by Generalized Lambert W.

for real $a,b,c,x$:

$$ax^{x-b}-c=0$$ $$ae^{\ln(x)(x-b)}-c=0$$ $$ae^{\ln(x)(x-b)}=c$$ $$e^{\ln(x)(x-b)}=\frac{c}{a}$$ $$\ln(e^{\ln(x)(x-b)})=\ln\left(\frac{c}{a}\right)$$ $$\ln(x)(x-b)=\ln\left(\frac{c}{a}\right)$$ $x=e^t$: $$\ln(e^t)(e^t-b)=\ln\left(\frac{c}{a}\right)$$ $$t(e^t-b)=\ln\left(\frac{c}{a}\right)$$ $$e^t=\frac{b\ t+\ln\left(\frac{c}{a}\right)}{t}$$ $$\frac{t}{b\ t+\ln\left(\frac{c}{a}\right)}e^t=1$$ $$\frac{b\ t}{(b\ t+\ln\left(\frac{c}{a}\right))}e^t=b$$ $$\frac{t}{t+\frac{1}{b}\ln\left(\frac{c}{a}\right)}e^t=b$$ $$\frac{t}{t-\left(-\frac{1}{b}\ln\left(\frac{c}{a}\right)\right)}e^t=b$$ $$t=W\left(^{\ \ \ \ \ \ \ \ \ 0}_{-\frac{1}{b}\ln\left(\frac{c}{a}\right)};\ b\right)$$ $$x=e^{W\left(^{\ \ \ \ \ \ \ \ \ 0}_{-\frac{1}{b}\ln\left(\frac{c}{a}\right)};\ b\right)}$$

The inverse relation of your kind of equations is what Mezö et al. call $r$-Lambert function. They write:
"Depending on the parameter $r$, the $r$-Lambert function has one, two or three real branches and so the above equations can have one, two or three solutions"
$\ $

[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