# What is the inverse of e^2x/3+e^2x?

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### 2 Answers

Parantheses are your friends:

(1) If you meant: find the inverse of `y=e^(2x)/3+e^(2x)`

Write (e^(2x)/3)+e^(2x)

The typical way to solve for the inverse is to exchange x and y and solve the resulting equation for y.

`x=e^(2y)/3+e^(2y)`

`x=4/3e^(2y)`

`e^(2y)=3/4 x`

`2y=ln(3/4x)`

`y=1/2ln(3/4x)` is the inverse we seek.

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(2) If you meant `e^(2x)/(3+e^(2x))`

Write e^(2x)/(3+e^(2x))

Again exchange x and y and solve for y:

`y=e^(2x)/(3+e^(2x))`

`e^(2y)/(3+e^(2y))=x` Multiply the left side by `e^(-2y)/e^(-2y)` to get

`1/(3e^(-2y)+1)=x`

`3e^(-2y)+1=1/x`

`3e^(-2y)=1/x-1`

`e^(-2y)=1/3((1-x)/x)`

`-2y=ln((1-x)/(3x))`

`y=-1/2ln((1-x)/(3x))` which is the inverse.

Write e^(2x)/(3+e^(2x))

Again exchange x and y and solve for y:

`y=e^(2x)/(3+e^(2x))`

`e^(2y)/(3+e^(2y))=x` Multiply the left side by `e^(-2y)/e^(-2y)` to get

`1/(3e^(-2y)+1)=x`

`3e^(-2y)+1=1/x`

`3e^(-2y)=1/x-1`

`e^(-2y)=1/3((1-x)/x)`

`-2y=ln((1-x)/(3x))`

`y=-1/2ln((1-x)/(3x))` which is the inverse.