Do you know how I could compute the inverse function of the following exponential sentence?
$$y=\dfrac{e^x}{1+2e^x}$$
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Can you solve for $x$ in terms of $y$? – André Nicolas Dec 01 '12 at 21:43
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1Is it at all possible that you wanted $e^x$ rather than $e^2$ in the numerator? (If so, then my answer below would have to get changed.) – Michael Hardy Dec 01 '12 at 21:48
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this is my result $F^-1(x)=ln((x/1-2x))$ – Diego Pacheco Dec 01 '12 at 21:49
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Michael you're right, what I want is $e^x$ rather than $e^2$constant. – Diego Pacheco Dec 01 '12 at 21:53
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For the amended question, the answer is correct. – André Nicolas Dec 01 '12 at 21:55
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OK, I've added to my answer in view of the amended question. – Michael Hardy Dec 01 '12 at 22:02
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$$ \text{exponentiate} \longrightarrow \text{multiply by 2}\longrightarrow\text{add 1} \longrightarrow \text{reciprocal}\longrightarrow\text{multiply by }e^2 $$
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.)
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Sorry but I don't get how you go from $\frac{e^x}{1+2e^x}$ to $\frac{1}{e^{-x}+2}$ – MphLee May 20 '13 at 20:24
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1@MphLee : $\dfrac{e^x}{1+2e^x}$ $=\dfrac{e^x}{1+2e^x}\cdot\dfrac{e^{-x}}{e^{-x}}$. When you multiply the numerators you get $e^x\cdot e^{-x}=1$. When you multiply the denominators you get $(1+2e^x)(e^{-x})$ $=(e^{-x}) + (2e^x e^{-x})$. And $2e^x e^{-x}=2\cdot1$. – Michael Hardy May 20 '13 at 23:32
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$$y=\dfrac{e^2}{1+2e^x}$$ $$1+2e^x=e^2/y$$ $$2e^x=e^2/y-1$$ $$e^x=\frac{e^2/y-1}{2}$$ $$x=\ln{\frac{e^2/y-1}{2}}$$
Adi Dani
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