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I would like to solve this equation
$$\left(\frac{1}{50}\right)^x=\log_{\frac{1}{50}}x$$
I plot graphs on GeoGebra and I found that there are three intersections (i.e. solutions) from this equation. I found one solution and below are my steps.
$$\begin{aligned} 50^{-x}&=-\log_{50}x\\x&=50^{-50^{-x}}\\&=50^{-x}\\x(50^x)&=1\\x\ln50(e^{x\ln50})&=\ln50\\x\ln50&=W(\ln50)\\x&=\frac{W(\ln50)}{\ln50}\end{aligned}$$ which has been confirmed a root in geogebra.
However, I have no idea how to find the other two roots of this equation.
Also, I noticed for an equation $$a^x=\log_{a}x$$ can have either one root or three roots for a certain range of real values a.
$a=\frac{1}{50}$ is an example of having three roots.
But I am not sure how to find this range to have three roots.
Can anyone help? Thank you so much!

IV_
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2 Answers2

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We see, this equation is a polynomial equation of more than one algebraically independent monomials $(\left(\frac{1}{50}\right)^x,\log_\frac{1}{50}(x))$ and with no univariate factor. We therefore don't know how to solve the equation for $x$ by rearranging by applying only finite numbers of elementary functions (elementary operations) we can read from the equation.

To solve the equation in closed form, we therefore have no choice but to use Special functions. The equation isn't in a form to solve it in terms of Lambert W.

Your step $x=50^{-x}$ is wrong.

You could use numerical methods or power series methods.

see also Wolfram Alpha

IV_
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$\DeclareMathOperator BB \DeclareMathOperator WW$Is there a way to calculate the zeros of $f(z,w)= w-z^{(z^w)}$? has a special function solution, but it is not standard and there is no explicit representation of it. One can use

Convert $\frac1b\sum_{n=1}^\infty\frac{(b e^a)^n}{n!}B_{n-1}(an)$ to integral using $B_n(x)=\frac{n!}{2\pi i}\oint\frac{e^{x(e^t-1)}}{t^{n+1}}dt$

with the Bell polynomial for $0<a<e^{-e}$ when there are $3$ real roots since the smallest and largest roots cannot generally be solved by $\W(x)$:

  • Smallest root:

$$a^{a^x}=x\implies x=\frac1{\ln(a)}\sum_{n=1}^\infty\frac{(a\ln(a))^n}{n!}\B_{n-1}(\ln(a)n)=-\frac1{2\pi\ln(a)}\int_{-\pi}^\pi e^{it}\ln\left(1-\ln(a)e^{-i t}a^{e^{e^{it}}}\right)dt$$

with the series and integral demonstrated in the links

  • $2$nd smallest root:

$$a^{a^x}=x\implies x=-\frac{\W(-\ln(a))}{\ln(a)}$$

  • Largest root:

$$a^{a^x}=x\implies x=1+\frac1{\ln(a)}\sum_{n=1}^\infty\frac{(a\ln(a))^n}{nn!}\B_n(\ln(a)n)=1-\frac1{2\pi\ln(a)}\int_{-\pi}^\pi \ln\left(1-\ln(a)e^{-i t}a^{e^{e^{it}}}\right)dt $$

with the series and integral demonstrated in the links again.

The integrals are close to those from Cauchy’s integral formula, so they are not completely unique solutions. The series seem to converge more quickly for $a\to0$ and more slowly for $a\to e^{-e}$

Тyma Gaidash
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