Solutions to $\left( \frac{1+3x}{1+2x} \right)^{\frac{1+x}{x}}=2$? I am looking for all non-negative real solutions to
$$
  \left( \frac{1+3x}{1+2x} \right)^{\frac{1+x}{x}}=2.
$$
Numerically it seems that there is a unique solution $x \approx 0.4256$. Any ideas how to prove/find it?

This is probably nowhere near a full answer, but steps could potentially be useful.
Recognize that $\frac{1+x}{x}=\frac1{x}+1$.  Then, the equation can be rewritten as
\begin{align}
\bigg(\frac{1+3x}{1+2x}\bigg)^{\frac{1+x}{x}}=\bigg(\frac{1+3x}{1+2x}\bigg)^{\frac1{x}+1}&=2\\
\bigg(\frac{\frac1{x}+3}{\frac1{x}+2}\bigg)^{\frac1{x}+1}&=2
\end{align}
Let $A=\frac1{x}+2$.  Then,
\begin{align}
\bigg(\frac{A+1}{A}\bigg)^{A-1}&=2\\
\bigg(1+\frac1{A}\bigg)^{A-1}&=2\\
\bigg(1+\frac1{A}\bigg)^{A}&=2\bigg(1+\frac1{A}\bigg)
\end{align}
