How do you find the inverse of #y = e^x/(1 + 4 e^x)#?

Question

Sophie Adams · Accepted Answer

undefined To find the inverse of $ y = \frac{e^x}{1 + 4e^x} $, first, interchange the roles of $ x $ and $ y $, then solve for $ y $ in terms of $ x $.

1. Start with the original equation: $ y = \frac{e^x}{1 + 4e^x} $.
2. Swap $ x $ and $ y $: $ x = \frac{e^y}{1 + 4e^y} $.
3. Multiply both sides by $ (1 + 4e^y) $: $ x(1 + 4e^y) = e^y $.
4. Distribute and rearrange terms: $ x + 4xe^y = e^y $.
5. Move all terms involving $ e^y $ to one side: $ e^y - 4xe^y = x $.
6. Factor out $ e^y $: $ e^y(1 - 4x) = x $.
7. Solve for $ e^y $: $ e^y = \frac{x}{1 - 4x} $.
8. Take the natural logarithm of both sides: $ y = \ln\left(\frac{x}{1 - 4x}\right) $.

Therefore, the inverse of the function $ y = \frac{e^x}{1 + 4e^x} $ is $ y = \ln\left(\frac{x}{1 - 4x}\right) $.

James Allen · Answer

#x =\ln(\frac{y}{1-4y})#

This question would be a "solving for the inverse of a rational functions question" and you would follow the same standard procedure as you would for solving those equations.

First multiply both sides by #1+4e^x# : #y(1+4e^x) =e^x# #y+4e^xy - e^x = 0# #4e^xy - e^x = -y# , factor #e^x# #e^x(4y - 1) = -y# #e^x= \frac{-y}{4y - 1} = \frac{y}{1-4y} # #x = \ln(\frac{y}{1-4y})#