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?

Starting from Andrew Chin's answer and taking logarithms, we need to solve
$$(A-1) \log \left(1+\frac{1}{A}\right)=\log(2)$$
Building the simple $[1,1]$ Padé approximant around $A=4$ gives 
$$(A-1) \log \left(1+\frac{1}{A}\right)\sim\frac{3 \log \left(\frac{5}{4}\right)+\frac{ \left(18+800 \log
   ^2\left(\frac{5}{4}\right)-201 \log \left(\frac{5}{4}\right)\right)}{40 \left(20
   \log \left(\frac{5}{4}\right)-3\right)}(A-4) } {1+\frac{13 }{40 \left(20 \log \left(\frac{5}{4}\right)-3\right)}(A-4) }$$ Then one linear equation in $A$ the approximate solution of which being
$$A=4+\frac{40 \log \left(\frac{128}{125}\right) \left(20 \log
   \left(\frac{5}{4}\right)-3\right)}{18+\log \left(\frac{5}{4}\right) \left(800
   \log \left(\frac{5}{4}\right)-201\right)-13 \log (2)}\approx 4.349417$$ while the "exact" solution is                                    $\approx 4.349409$.
Then $x$. 
We could go further building, around $A=4$, the $[1,n]$ Padé approximant of
$$(A-1) \log \left(1+\frac{1}{A}\right)-\log(2)$$ Setting the numerator equal to $0$ would give exact (but nasty) expressions. The numerical values are reported below
$$\left(
\begin{array}{cc}
 n & A_{(n)} \\
 0 &  4.324246310 \\
 1 &  4.349417080 \\
 2 &  4.349408445 \\
 3 &  4.349409126 \\
 4 &  4.349409072 \\
 5 &  4.349409076 
\end{array}
\right)$$
Edit
It could be interesting to notice that, assuming that $A$ is large
$$(A-1) \log \left(1+\frac{1}{A}\right)=1-\frac{3}{2 A}+\frac{5}{6 A^2}+O\left(\frac{1}{A^3}\right)$$ which gives as an estimate
$$A=\frac{9+\sqrt{3 (40 \log (2)-13)}}{12 (1-\log (2))} \approx 4.24922$$
