Do you know how I could compute the inverse function of the following exponential sentence?
$y=\dfrac{e^x}{1+2e^x}$
Do you know how I could compute the inverse function of the following exponential sentence?
$y=\dfrac{e^x}{1+2e^x}$
$ \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.)
$y=\dfrac{e^2}{1+2e^x}$ $1+2e^x=e^2/y$ $2e^x=e^2/y-1$ $e^x=\frac{e^2/y-1}{2}$ $x=\ln{\frac{e^2/y-1}{2}}$