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?
 A: 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}
A: Here's a graph:

Numerical methods give:  $x \approx 0.425639$ as the only solution.
I don't see any formal, algebraic routes to a solution, including Lambert's W function.
A: 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
$$A=\frac{9+\sqrt{3 (40 \log (2)-13)}}{12 (1-\log (2))} \approx 4.24922$$
A: Let $u = \large{1+3x \over 1+2x}$, we have
$$\large f(u) = u^{2-u \over u-1} - 2 = 0$$
$x > 0 →\quad 1 < u < 1.5$
Plot of f(u) for this full range indicated 1 and only 1 root
Solving with secant's method :
$\begin{matrix}
u & f(u) \cr
1.20 & +0.0736 \cr
1.25 & -0.046875 \cr
1.230545756 & -0.00149914371 \cr
1.229903020 & +3.162582158·10^{-5} \cr
1.229916299 & -2.087064255·10^{-8} \cr
1.229916290
\end{matrix}$
$$x = {1-u \over 2u-3} ≈ 0.4256389448$$
