I'm really not doing well in Calculus. Can anyone help me with finding the inverse function of e^x / 1+ 5e^x?

Question

It seems so simple! But my Calc professor lectures quite the opposite that my Pre-Calc professor did. Any help would be greatly appreciated!

Answer 1

e^x / (1+5e^x) = y

e^x = y(1+5e^x)

e^x = y + 5ye^x

1 = y / e^x + 5y

1 = ye^-x + 5y

0 = ln (ye^-x+5y)

0 = ln (y(e^-x+5))

0 = ln y + ln (e^-x + 5)

ln 1/y = ln (e^-x+5)

1/y = e^-x + 5

1/y - 5 = e^-x

ln (1/y - 5) = -x

ln (1/y - 5)^-1 = x

ln (1/x - 5)^-1 = y

FYI, when the fraction in the natural log is simplified... it's the same as the bottom two answers.

Answer 2

Sure thing. I'll walk you through it:

You start with this expression:
y=e^x/(1+5e^x)

switch the x and y, and then solve for y:
x=e^y/(1+5e^y)

Multiply both sides by (1+e^y):
x(1+5e^y)=e^y

Distribute the x:
x+5xe^y=e^y

Subtract (5xe^y) from both sides:
x=e^y-5xe^y

Factor right side by e^y:
x=e^y(1-5x)

Divide both sides by (1-5x):
x/(1-5x)=e^y

Now take the ln of both sides:
ln(x/(1-5x))=ln(e^y)

Since ln(e^y)=y:
ln(x/(1-5x))=y

And that is your inverse function. To sum up, switch your x and y variables and then just solve for y. Good luck!

Answer 3

y = e^x/(1+5e^x)
y+5ye^x = e^x
y = e^x -5ye^x
y = e^x(1-5y)
y/(1-5y) = e^x
ln[y/(1-5y)] = ln e^x = xln e = x
If you interchange x and y , you get :
y = ln[x/(1-5x)]
You must check this because not all functions have an inverse.

In this case the inverses do not work.