How do you find the inverse of #y = (e^x)/(1+2e^x)#?

Question

Christian Antunez · Accepted Answer

undefined To find the inverse of the function $y = \frac{e^x}{1 + 2e^x}$, follow these steps:

1. Replace $y$ with $x$ and $x$ with $y$:
$$x = \frac{e^y}{1 + 2e^y}$$

2. Solve this equation for $y$:
$$x(1 + 2e^y) = e^y$$

3. Expand and rearrange terms:
$$x + 2xe^y = e^y$$

4. Move terms involving $e^y$ to one side:
$$2xe^y - e^y = -x$$

5. Factor out $e^y$:
$$e^y(2x - 1) = -x$$

6. Divide both sides by $2x - 1$:
$$e^y = \frac{-x}{2x - 1}$$

7. Take the natural logarithm of both sides to isolate $y$:
$$y = \ln\left(\frac{-x}{2x - 1}\right)$$

Therefore, the inverse of the function $y = \frac{e^x}{1 + 2e^x}$ is given by $y = \ln\left(\frac{-x}{2x - 1}\right)$.

Anna Allen · Answer

Step by step working is shown below.

Interesting problem! Finding inverse of #y=e^x/(1+2e^x)# For inverse understand that the graph would be a reflection over the #x=y# line. So the starting point would be to understand #(x,y)->(y,x)# when finding the inverse Step 1: Swap #x# and #y# #x = e^y/(1+2e^y)# Step 2: Solve for #y# We would start by multiplying both sides with #(1+2e^y)#. This is done to remove the denominator. #x(1+2e^y) = e^y# Use distributive property. #x+2xe^y=e^y# Collect all terms containing #e^y# to one side of the equation. Subtracting both sides by #2xe^y# should do the trick. #x = e^y-2xe^y# Factor out #e^y# from the right side of the equation. This is the reverse process of distribution. #x = e^y(1-2x)# We are solving for #y# and for that we need #e^y# isolated. To remove #(1-2x)# we would divide both sides by #(1-2x)#. #x/(1-2x) = e^y# To solve for #y# we need to convert this equation to natural logarithmic equation. Take #ln# on both sides. Note: #ln(e^a) = a# #ln(x/(1-2x)) = y# #y=ln(x/(1-2x))# This is the equation of the inverse.