# Solving for $x$ in terms of $y$ for equations involving quadratics, logarithms, and exponentials.

For the equation:

$$y^2 = \frac{\ln(1 - xe^{xy^2})}{1 - xe^{xy^2}}$$

How might one go about solving for $$x$$ in terms of $$y$$? First, I attempted solving the equation by first performing the subsitution $$u = xe^{xy^2}$$, which then reduces the problem to: $$\frac{\ln(\frac{u}{x})}{x} = \frac{\ln(1 - u)}{1-u}$$

Unfortunately, I could not find where to go from there. Secondly, I tried multiplying both sides of the initial equation by $$1 - xe^{xy^2}$$ to obtain:

$$y^2-xy^2e^{xy^2} = \ln(1 - xe^{xy^2})$$

From here, I could subtract $$y^2$$ from both sides and divide by -1 to get the equation in a form where the product logarithm could be applied, but unfortunately the right hand side would contain both $$x$$ and $$y$$. Exponentiating both sides with base $$e$$ only seems to make the problem more convoluted and messy than it already is. So how might I go about actually solving this equation for $$x$$?

Considering $$y^2 = \frac{\ln(1 - xe^{xy^2})}{1 - xe^{xy^2}}$$ we could start defining $$z=1 - xe^{xy^2}\implies y^2=\frac{\log(z)}z$$ Solving for $$z$$ $$z=-\frac{W\left(-y^2\right)}{y^2}$$ where $$W(.)$$ is Lambert function. So $$1 - xe^{xy^2}=-\frac{W\left(-y^2\right)}{y^2}$$ and, again, Lambert function $$x=\frac{W\left(y^2+W\left(-y^2\right)\right)}{y^2}$$

Do not ask me for $$y(x)$$, please !

Edit

Since you mention the product logarithm, I suppose that you can run Mathematica. If so,ask for the contour plot of $$f(x,y)=y^2 - \frac{\ln(1 - xe^{xy^2})}{1 - xe^{xy^2}}=0$$ for $$-\frac{1}{\sqrt{e}} \leq y \leq \frac{1}{\sqrt{e}}$$ and $$-1000 \leq x \leq 0$$.

The substitution: $$u=x\;\exp \left( x~y^{2}\right) =x\;\exp \left( x~v\right) ;\;v=y^{2}$$ is the the correct Ansatz here. This leads to: $$\frac{\log \left( \frac{u}{x}\right) }{x}=\frac{\log \left( 1-u\right) }{1-u}$$ Application of the functional equation on the left hand side of the equation leads to: $$\frac{\log \left( u\right) }{x}-\frac{\log \left( x\right) }{x}=\frac{\log \left( 1-u\right) }{1-u}$$ Now we substitute $$u$$ on the left side of the equation:

$$\frac{\log \left( x\;\exp \left( x~v\right) \right) }{x}-\frac{\log \left( x\right) }{x}=\left( v+\frac{\log \left( x\right) }{x}\right) -\frac{\log \left( x\right) }{x}=v$$ and recognize, that the term: $$\frac{\log \left( x\right) }{x}$$ vanishes. Now, we can solve for $$u$$ $$u=\frac{v+W\left( -v\right) }{v}$$ where $$W\left( z\right)$$ is the Lambert function [Mathematica] (http://functions.wolfram.com/ElementaryFunctions/ProductLog/27/02/) or [Corless] (https://cs.uwaterloo.ca/research/tr/1993/03/W.pdf). In the end we have to substitute $$u\rightarrow x\;\exp \left( x~v\right)$$ and $$v$$ and solve the equation for $$x$$ $$x=\frac{W\left( y^{2}+W\left( -y^{2}\right) \right) }{y^{2}}$$ Numerical comparisons for $$y=0.316228$$ e.g. $$v=0.1$$ confirms the result.

• So I was a bit to late! – stocha Feb 18 '19 at 11:31