Simple way to see that $r=\frac{st}{s+t}$?

Question

Consider the system of equations

$$ f(x)=g(x)=h(x) $$

where $f(x)= \exp\bigg(\frac{s}{\log x} \bigg) $ and $g(x)=\exp\bigg( \frac{t}{\log(1-x)} \bigg)$ and $h(x)=\exp\bigg(\frac{r}{\log x}+\frac{r}{\log (1-x)}\bigg)$. Here $s,t\in \Bbb N$ and $r\in \Bbb Q_+$ and $x\in (0,1),$ and where $r=\frac{st}{s+t}$.

Is there a simple way to see that $r=\frac{st}{s+t}$?

I manipulated the equations but I still don't see why $r$ should be related to $s$ and $t$ like that.

Answer 1

Since $\exp$ is injective, the equality $f(x)=g(x)=h(x)$ implies that $$\frac{s}{\log x} = \frac{t}{\log (1-x)} = r \left(\frac{1}{\log x} + \frac{1}{\log(1-x)}\right)$$ for all $x \in (0,1).$ The idea now is to pick a particular $x$. In this setting $x = 1/2$ seems natural due to the symmetry. Plugging $x=1/2$ in the last equation leads to $$\frac{s}{\log(1/2)} = \frac{t}{\log (1/2)} = \frac{2r}{\ln(1/2)}.$$ Hence $$s=t=2r.$$ From these equalities, you can easily deduce that $$\frac{st}{s+t} = \frac{(2r)(2r)}{2r+2r}=\frac{4r^2}{4r}=r.$$

Answer 2

You can simplify $f(x)=g(x)$ to $$\frac{s}{\log x} = \frac{t}{\log (1-x)}$$ which leads to $$\frac{\log (1-x)}{\log x} = \frac{t}{s}$$

Similarly you can simplify $f(x)=h(x)$ to: $$\frac{s}{\log x} = r \left(\frac{1}{\log x} + \frac{1}{\log(1-x)}\right)\\ \frac{s-r}{\log x} = \frac{r}{\log(1-x)} \\ \frac{\log(1-x)}{\log x} = \frac{r}{s-r} $$

These two equations have the same left hand sides, so their right hand sides are equal too, giving: $$ \frac{t}{s} = \frac{r}{s-r} $$ Solving for $r$ gives: $$ t(s-r) = rs \\ st-rt=rs \\ st = r(s+t) \\ r=\frac{st}{s+t}$$