algebra-precalculus
I'm trying to solve for $y$ in terms of $x$ for the expression below.
$$\frac{\ln(x)\ln(y)}{\ln(1-x)\ln(1-y)}=1$$
First I multiplied both sides by $$ \frac{\ln(1-x)}{\ln(x)} $$
to get
$$ \frac{\ln(y)}{\ln(1-y)}=\frac{\ln(1-x)}{\ln(x)} $$
but I don't see how to isolate $y.$ I tried using every technique I know including logarithm properties.
As percusse and GEdgar point-out in there comments that the reason this seemingly simple equation is not solvable using simple algebra lies in the fact that that the LHS of $2^x = x^2$ is a transcendental function. i.e. it cannot be expressed as a polynomial. Actually the closest it can come to in a "polynomial" form is its Maclaurin series form (see below).
Using pre-calculus techniques you can show, for instance, that you can take log of both sides as in
$$2^x = x^2$$ $$\implies ln(2^x) = ln(x^2) \quad \forall x \ne 0 $$ $$\implies x ln(2) = 2 ln(x) $$ $$\implies ln(x) = {2x \over ln(2)} \quad \textbf {(A)}$$
So the solution to our problem are all values of $x$ that are the roots of equation $\textbf{(A)}$ ... Pre-calculus you can use graphing techniques to determine the answer.
Solving transcendental functions, in general, requires a lot of different calculus techniques, that are probably beyond the scope of this answer.
Using Taylor's Theorem (which is part of calculus) we can show that:
$$e^u = \sum_{n=0}^{ \infty } {u^n \over n!} = 1 + {u^1 \over 1!} + {u^2 \over 2!} + {u^3 \over 3!} + {u^4 \over 4!} + \cdots \quad \textbf{(B)}$$
For considerable historical reasons $\textbf{(B)}$ is called Maclaurin series for $e^u$. You can find Maclaurin series for a large number of functions that have certain properties.
For purpose of this discussion, assume that (B) is provable. We can use it to express the infinite series for $2^x$ by noting that $2 \equiv e^{ln(2)}$, and that $(a^x)^y = (a)^{xy}$.
$$ [2]^x = [e^{ln(2)}]^x = [e^{ln(2) \dot x}]$$
Substituting $u$ with $2^x$ in $\textbf{(B)}$ power-series we get:
$$2^x = ln(2) \sum_{n=0}^{ \infty } {x^n \over n!} = ln(2) \left( 1 + {x^1 \over 1!} + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots \right)$$
As you can see solving (A) without knowing some more properties about behavior of $e^x$ becomes intractable. That is what calculus is all about ;) once you get into it, you wil see that these problems become solvable. Although, the solutions are, by no means, trivial.
$$e^{\frac{x^2}{4vt}} = 1+\frac{x^2}{2vt}$$ Let $y=\frac{x^2}{4vt}$ $$e^y=1+2y$$ $$e^{-y}=\frac{1}{1+2y}$$ $$(1+2y)e^{-y}=1$$ $$(\frac12+y)e^{-y}=\frac12$$ $$(-\frac12-y)e^{-y}=-\frac12$$ $$ (-\frac12-y)e^{-y}e^{-\frac12}=-\frac12 e^{-\frac12}$$ $$ (-\frac12-y)e^{-\frac12-y}=-\frac12 e^{-\frac12}=-\frac{1}{2\sqrt{e}}$$ $X=(-\frac12-y)$ $$Xe^X=-\frac{1}{2\sqrt{e}}$$ From the definition of the Lambert W function : http://mathworld.wolfram.com/LambertW-Function.html $$X=W\left(-\frac{1}{2\sqrt{e}}\right)$$ $$y=-\frac12-X=-\frac12-W\left(-\frac{1}{2\sqrt{e}}\right)$$ $$x=\sqrt{4vty}=\sqrt{-2-4W\left(-\frac{1}{2\sqrt{e}}\right)}\sqrt{vt}$$ The Lambert W(z) function is a multi valuated function in $-\frac{1}{e} <z<0$ , real $z$.
This is presently the case where $z=-\frac{1}{2\sqrt{e}}$ since $-\frac{1}{e} <-\frac{1}{2\sqrt{e}}<0$
First root :
$W_0\left(-\frac{1}{2\sqrt{e}}\right)=-\frac12 \quad;\quad {-2-4W_0\left(-\frac{1}{2\sqrt{e}}\right)}=-2-4(-1/2)=0 \quad;\quad x=0$
Second root :
$W_{-1}\left(-\frac{1}{2\sqrt{e}}\right)\simeq -1.756431... $
https://www.wolframalpha.com/input/?i=lambertw(-1,-1%2F(2+sqrt(e)))
One can use series expansion to compute it approximately. See page 13 in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function . The convergence is very slow. See the numerical calculus in the below addition. In practice, it is certainly faster to solve directly $e^y=1+2y$ with Newton-Raphson or similar iterative method.
Finally, an approximate value is :
$$x\simeq\sqrt{-2-4(-1.756431)}\sqrt{vt}\simeq 2.2418128 \sqrt{vt}$$
With more digits : https://www.wolframalpha.com/input/?i=sqrt(-2-4+lambertw(-1,-1%2F(2+sqrt(e))))
IN ADDITION :
Example of recursive numerical calculus of $W_{-1}(x)$ in the range $-\frac{1}{e}<x<0$
Best Answer
The function $f(x)=\dfrac{\ln(1-x)}{\ln(x)}$ is monotonic in its domain $(0,1)$, hence it is invertible. So the relation between $x$ and $y$ is a bijection, and…
$$y=1-x.$$
Interestingly, the function is well approximated by $\left(\dfrac1x-1\right)^{-3/2}$, and a solution with $a$ in the RHS is approximately
$$\left(\dfrac1x-1\right)^{-3/2}=a\left(\dfrac1y-1\right)^{3/2},$$ or
$$y=\frac{1-x}{1+(a^{2/3}-1)x}.$$