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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... enter image description here

Can someone help me to solve for $x$ and explain this transformation that W.A. gave me

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  • $\begingroup$ Don't let the $e^x$ rattle you. Replace $e^x$ with $m$ and solve $\frac m{m+2} = .9$ ($m = .9(m+2)=.9m +1.8;.1*m = 1.8; m =e^x = 18$.) Once you solve $e^x = m$ take the natural log and get $x = \ln m$. (So $m=e^x = 18$ so $x = \ln m = \ln 18$) $\endgroup$ – fleablood Nov 11 '17 at 23:17
  • $\begingroup$ "I tried step by step solution in Wolfram Alpha" Wolfram Alpha is a brainless computer algorithm. It is not a human being. $\frac {e^x}{e^x + 2} = \frac {e^x + 2 - 2}{e^x+2} = 1 - \frac 2{e^x +2}$. Let $y=\frac 1{e^x+2}$ so $1-2y =.9; y=.05$ so $\frac 1y = 20$. So $e^x+2 = 20$ so $e^x = 18$. So $x = \ln 18$. It make sense to program a computer to do this but for you or me.... what a lot of wasted steps and effort. $\endgroup$ – fleablood Nov 11 '17 at 23:23
  • $\begingroup$ So how would you tell an automaton how to solve $\frac {f(x)}{g(x)}=h(x)$? Well, you'd tell a human being to look at $f(x) = h(x)g(x)$ and see if it looks solvable and if it doesn't try something else. Well, you tell an automaton to reduces $\frac{f(x)}{g(x)}=\frac{f(x)=kg(x) + g'(x)}{g(x)} = k + \frac {g'(x)}{g(x)}=h(x)$ and keep repeating until you get something simple. That will eventually fall to bits but... it's a lot of human work to get to that stage and instead there is very likely something a lot simpler. $\endgroup$ – fleablood Nov 11 '17 at 23:44
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Presumably, $y$ is just a variable introduced for simplification purposes. Try approaching it this way. $$ \frac{e^x}{e^x+2}=0.9$$ $$ e^x=0.9(e^x+2)$$ $$ e^x=0.9e^x+1.8$$ $$ (1-0.9)e^x=1.8$$ $$ 0.1e^x=1.8$$ $$ e^x=18$$ $$ x=\log 18.$$

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  • $\begingroup$ yeah that's really easy way of doing it. how I did not see it lol $\endgroup$ – YohanRoth Nov 11 '17 at 22:56
  • $\begingroup$ Happens to the best of us :-) $\endgroup$ – Alekos Robotis Nov 11 '17 at 22:57
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$$1-\frac{2}{e^x+2}=\frac{1(e^x+2)-2}{e^x+2}=\frac{e^x}{e^x+2}$$ and so $$\frac{e^x}{e^x+2}=0.9$$ $$\implies 1-\frac{2}{e^x+2}=0.9$$ $$\implies \frac{2}{e^x+2}=1-0.9=0.1$$ $$\implies e^x+2=\frac{2}{0.1}=20$$ $$\implies e^x=18$$ $$\implies x=\ln(18)$$

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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 $$x=\ln{18}$$

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$$\frac{e^x}{e^x+2} = \frac{e^x + 2}{e^x+2} - \frac{2}{e^x+2} = 1 - \frac{2}{e^x+2}.$$ After setting this equal to $0.9$, it should be easier to solve for $x$ now.

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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 $u = e^x, v = 0.9, A=1, B=0, C=1,D=2.$ $$ \left( \begin{array}{cc} A & B \\ C & D \\ \end{array} \right) \left( \begin{array}{cc} D & -B \\ -C & A \\ \end{array} \right) = \; \; (AD-BC) \; \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) $$

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