Do you know how I could compute the inverse function of the following exponential sentence?
$$y=\dfrac{e^x}{1+2e^x}$$
Best Answer
$$
\text{exponentiate} \longrightarrow \text{multiply by 2}\longrightarrow\text{add 1} \longrightarrow \text{reciprocal}\longrightarrow\text{multiply by }e^2
$$
What gets done last gets undone first.
So the inverse is:
$$
\text{Divide by }e^2\longrightarrow\text{reciprocal} \longrightarrow\text{subtract 1} \longrightarrow \text{divide by 2}\longrightarrow\text{take logarithm}
$$
I.e. if $y = \dfrac{e^2}{1+2e^x}$ then $x = \log_e\left(\dfrac{\dfrac{1}{y/e^2} - 1}{2}\right)$. (Then simplify.)
LATER EDIT:
Since you've now said you wanted $e^x$ in the numerator, here's that:
$$
y=\frac{e^x}{1+2e^x} = \frac{1}{e^{-x}+2}
$$
$$
\text{multiply by }-1 \longrightarrow \text{exponentiate} \longrightarrow \text{add 2} \longrightarrow\text{reciprocal}
$$
So the inverse is:
$$
\text{reciprocal} \longrightarrow\text{subtract 2}\longrightarrow\text{take logarithm} \longrightarrow\text{multiply by }-1
$$
(Two of the steps are their own inverses.) So
$$
x = -\log_e\left(\frac1y-2\right)
$$
(Then simplify.)
This function probably won't have a nice inverse, since finding the inverse is equivalent to solving the equation $f(x)=c$, or equivalently finding the roots of the equation $x^5+x^3+x-c=0$. This is a quintic polynomial, and as such probably will likely not have a general solution (i.e. for all $c$) in terms of radicals. The best you can do is usually some kind of numerical method.
Best Answer
$$ \text{exponentiate} \longrightarrow \text{multiply by 2}\longrightarrow\text{add 1} \longrightarrow \text{reciprocal}\longrightarrow\text{multiply by }e^2 $$
What gets done last gets undone first.
So the inverse is: $$ \text{Divide by }e^2\longrightarrow\text{reciprocal} \longrightarrow\text{subtract 1} \longrightarrow \text{divide by 2}\longrightarrow\text{take logarithm} $$
I.e. if $y = \dfrac{e^2}{1+2e^x}$ then $x = \log_e\left(\dfrac{\dfrac{1}{y/e^2} - 1}{2}\right)$. (Then simplify.)
LATER EDIT:
Since you've now said you wanted $e^x$ in the numerator, here's that: $$ y=\frac{e^x}{1+2e^x} = \frac{1}{e^{-x}+2} $$ $$ \text{multiply by }-1 \longrightarrow \text{exponentiate} \longrightarrow \text{add 2} \longrightarrow\text{reciprocal} $$
So the inverse is: $$ \text{reciprocal} \longrightarrow\text{subtract 2}\longrightarrow\text{take logarithm} \longrightarrow\text{multiply by }-1 $$
(Two of the steps are their own inverses.) So $$ x = -\log_e\left(\frac1y-2\right) $$ (Then simplify.)