Solving an equation analytically of the type $n \cdot x^{n - 1} \cdot y + \frac{x^n}{z} = w$ for n

Question

I am currently faced with solving the following equation for my thesis. I am looking for a symbolic answer to n $$ k \left( n \cdot d_{\text{rz}}^{n-1} \cdot \frac{\partial d_{\text{rz}}}{\partial r} + \frac{d_{\text{rz}}^n}{r} \right) = \rho \left( \frac{\partial v_z}{\partial r} \cdot v_r + \frac{\partial v_z}{\partial z} \cdot v_z \right)$$ So this equation can be written for simplification purposes as: $$n \cdot x^{n - 1} \cdot y + \frac{x^n}{z} = w$$ Now I would like to solve this equation analytically for n. Ideas that I have are trying to extract n by:

1. The the logarithm of both sides and apply various rules for mathematical operations using logarithms
2. Use the Lambert function and attempt to write the equation in an adequate form respectively I have namely tried both methods, unfortunately to no avail.

I have previously used the Lambert function to extract n from $n \cdot x^{n - 1} = y$. This has worked but applying this methodology to my case, namely $n \cdot x^{n - 1} \cdot y + \frac{x^n}{z} = w$, will not work (I believe).

Any guidance will be much appreciated as I would like to come to a symbolic answer for n!

Thanks in advance,

P.S. All numbers should be real, n should belong to [0,1]

Answer 1

more detailed:

$$nx^{n-1}y+\frac{x^n}{z}=w$$ $$\frac{nye^{n\ln(x)}}{x}+\frac{e^{n\ln(x)}}{z}=w$$ $$\left(\frac{y}{x}n+\frac{1}{z}\right)e^{n\ln(x)}=w$$ $$\frac{x}{y}\ln(x)\left(\frac{y}{x}n+\frac{1}{z}\right)e^{n\ln(x)}=\frac{x}{y}\ln(x)w$$ $$(n\ln(x)+\frac{x}{yz}\ln(x))e^{n\ln(x)}=\frac{x}{y}\ln(x)w$$ $$(n\ln(x)+\frac{x}{yz}\ln(x))e^{n\ln(x)+\frac{x}{yz}\ln(x)}=\frac{x}{y}\ln(x)we^{\frac{x}{yz}\ln(x)}$$ $$n\ln(x)+\frac{x}{yz}\ln(x)=W\left(\frac{x}{y}\ln(x)we^{\frac{x}{yz}\ln(x)}\right)$$ $$n=\frac{1}{\ln(x)}W\left(\frac{x}{y}\ln(x)we^{\frac{x}{yz}\ln(x)}\right)-\frac{x}{yz}$$ $$n=\frac{1}{\ln(x)}W\left(\frac{x}{y}\ln(x)wx^\frac{x}{yz}\right)-\frac{x}{yz}$$

Answer 2

$$ nyx^{n-1} + \frac{x^n}{z} =w $$$$ nyx^{-1} + \frac{1}{z} =wx^{-n}$$ We can set $\psi := yx^{-1}$ and $q= \frac{1}{z}$ $$ \psi n+ q= wx^{-n} $$ $$ \psi(n+ \frac{q}{\psi})= wx^{-n} $$ Now let $u := n+\frac{q}{\psi} \implies n =u-\frac{q}{\psi}$ $$ \psi u = wx^{-u+\frac{q}{\psi}} $$$$ u = \frac{w}{\psi}x^{-u}x^{\frac{q}{\psi}} $$ Set $\eta = \frac{w}{\psi}x^{\frac{q}{\psi}}$

So we simplify to: $$ u = \eta x^{-u} $$ $$ x^{u}u = \eta\implies e^{u\ln x} u \ln x = \eta \ln x $$ So we have at the end $$ u = \frac{1}{\ln x}\operatorname{W_n}(\eta \ln x) $$