How to solve for $x$ in $\frac{e^x}{e^x + 2} = 0.9$ How to solve for $x$ in $\frac{e^x}{e^x + 2} = 0.9$ ?
I took log on both sides and arrived to 
$$x - \log(e^x + 2) = \log0.9$$
and I am not sure how to proceed...
I tried step by step solution in Wolfram Alpha, but I do not understand how they got the following step...

Can someone help me to solve for $x$ and explain this transformation that W.A. gave me
 A: Presumably, $y$ is just a variable introduced for simplification purposes. Try approaching it this way.
$$ \frac{e^x}{e^x+2}=0.9$$
$$ e^x=0.9(e^x+2)$$
$$ e^x=0.9e^x+1.8$$
$$ (1-0.9)e^x=1.8$$
$$ 0.1e^x=1.8$$
$$ e^x=18$$
$$ x=\log 18.$$
A: $$1-\frac{2}{e^x+2}=\frac{1(e^x+2)-2}{e^x+2}=\frac{e^x}{e^x+2}$$
and so
$$\frac{e^x}{e^x+2}=0.9$$
$$\implies 1-\frac{2}{e^x+2}=0.9$$
$$\implies \frac{2}{e^x+2}=1-0.9=0.1$$
$$\implies e^x+2=\frac{2}{0.1}=20$$
$$\implies e^x=18$$
$$\implies x=\ln(18)$$
A: Try comparing reciprocals of both sides:
$$\frac{e^x+2}{e^2} = \frac {10}9$$
leads to
$$1+\frac 2{e^x}=\frac{10}9$$
hence
$$\frac 2{e^x}=1/9$$
and
$$2\cdot 9=e^x$$
so
$$x=\ln{18}$$
A: $$\frac{e^x}{e^x+2} = \frac{e^x + 2}{e^x+2} - \frac{2}{e^x+2} = 1 - \frac{2}{e^x+2}.$$
After setting this equal to $0.9$, it should be easier to solve for $x$ now.
A: I don't see any mention of the Moebius transformation here. If we have quantities $u,v,$ with
$$  \frac{Au + B}{C u + D} = v,   $$ and $AD - BC \neq 0,$ then
$$ u = \frac{ \; Dv - B}{ \; \; -C v + A} $$
In your case $u = e^x, v = 0.9, A=1, B=0, C=1,D=2.$
$$
\left(
\begin{array}{cc}
A & B \\
C & D \\
\end{array}
\right)
\left(
\begin{array}{cc}
D & -B \\
-C & A \\
\end{array}
\right) = \; \;  (AD-BC) \;
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right)
$$
