How do I get an analytical solution to this nonlinear equation?

Question

I posted this question over on Math Stack Exchange (link), but have not received a response. I'm wondering if it's too complicated for that audience, so I'm posting it here in the hopes that someone here may be able to help me more.

Equations (3(a)-(b)) and (4(a)-(b)) from "Numerical Experiments on Application of Richardson Extrapolation With Nonuniform Grids" (DOI) provide the following solution to a nonlinear equation:

My question: How do they get from (2) to (3)?

To provide a little more detail, for various reasons (that aren't really relevant to the question), I am using the following expressions for $a_{1}$, $a_{2}$, and $a_{3}$:

$$ \begin{matrix} a_{1}=1 \\ a_{2}=r_{2}^p \\ a_{3}=r_{3}^{2p} \end{matrix} $$

Additionally, instead of the dependent variable being $F$, I'm using $w$. When the dependent variable and the constant ($C$) are eliminated from equations (2(a)-(c)) the following nonlinear equation is derived that needs to be solved for $p$:

$$ \frac{w_{3}-w_{2}}{w_{2}-w_{1}}=r_{3}^p \frac{\left (\frac{r_{2}}{r_{3}} \right )^p-r_{3}^p}{1-r_{2}^p} \tag{2.1}\label{eq21} $$

Unless I've made a mistake somewhere, using the notation I have above, I believe equations (3(a)-(b)) and (4(a)-(b)) become the following:

$$ p=\left | \frac{\ln \left |\frac{w_{2}-w_{3}}{w_{1}-w_{2}} \right |}{\ln \left (r_{2} \right )}+q\left ( p \right ) \right | \tag{3.1a}\label{eq31a} $$ $$ q(p)=\frac{\ln\left (\frac{\left (\frac{r_{3}^2}{r_{2}} \right )^p-s}{r_{2}^p-s} \right )}{\ln \left ( r_{2} \right )} \tag{3.1b}\label{eq31b} $$ $$ s=1 \cdot \text{sgn} \left ( \frac{w_{3}-w_{2}}{w_{2}-w_{1}} \right ) \tag{3.1c}\label{eq31c} $$

In short, How do I get from \eqref{eq21} to \eqref{eq31a}, \eqref{eq31b}, and \eqref{eq31c}?

EDIT:

I just discovered this (link) paper which identifies the type of equation I'm looking at here as a "transcendental equation" (see equation 10).

Answer 1

Taking the ratio of 2b) and 2a) resp. 2c) and 2b) and taking the logarithm yields \begin{eqnarray} \ln\left(\frac{F-F_1}{F-F_2}\right)&=&n\cdot \ln\left(\frac{a_1}{a_2}\right)\\ \ln\left(\frac{F-F_2}{F-F_3}\right)&=&n\cdot \ln\left(\frac{a_2}{a_3}\right) \end{eqnarray} Taking again the ratio yields $$ \ln\left(\frac{F-F_1}{F-F_2}\right)ln\left(\frac{a_2}{a_3}\right)=\ln\left(\frac{F-F_2}{F-F_3}\right)\ln\left(\frac{a_1}{a_2}\right) $$ It follows that $$ (F-F_1)^{\ln(a_2)}(F-F_2)^{\ln(a_3)}(F-F_3)^{\ln(a_1)}=(F-F_1)^{\ln(a_3)}(F-F_2)^{\ln(a_1)}(F-F_3)^{\ln(a_2)}, $$ i.e. an equation with the only unknown $F$. If this is solved we can solve for $n$ using $$ n=\frac{\ln\left(\frac{F-F_1}{F-F_2}\right)}{\ln\left(\frac{a_1}{a_2}\right)}=\frac{\ln(F-F_1)-\ln(F-F_2)}{\ln(a_1)-\ln(a_2)}. $$ Then one can solve for $C$ using one of the equations 2a),2b) resp. 2c).