# Thread: Find the Inverse?

1. ## Find the Inverse?

Hello, I've been trying to figure out how to solve this function but I keep getting stuck. The question asks:

Find a formula for the inverse of the function:

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

My teacher said to substitute the variable p to replace e^x to make the function easier to solve.

y = p / 1+5p^2

I'm not really sure how to go about solving this question. Would I multiply all sides by the denominator to get rid of it and then somehow solve from there?

Help is appreciated, so thanks in advance.

2. Originally Posted by dark-ryder341

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

My teacher said to substitute the variable p to replace e^x to make the function easier to solve.

y = p / 1+5p^2
How did you arrive at $p^2$ ?

What is the original function?

$y = \frac{e^x}{ 1+5e^x}$

or

$y = \frac{e^x}{ 1}+5e^x$

3. Oops, my mistake. I meant to write y = p / 1+5p

The p is just a substitution for e^x

The original function is:

Thanks.

4. Originally Posted by dark-ryder341
Oops, my mistake. I meant to write y = p / 1+5p

The p is just a substitution for e^x

The original function is:

Thanks.
The inverse is given by $y = f^{-1}(x)$ where $x = \frac{e^y}{1 + 5 e^y}$.

Your job is to make y the subject. Let $p = e^y$:

$x = \frac{p}{1 + 5p} \Rightarrow x (1 + 5p) = p \Rightarrow x + 5px = p \Rightarrow x = p - 5px$

$\Rightarrow x = p(1 - 5x) \Rightarrow p = \frac{x}{1 - 5x}$.

Substitute back for p and solve for y.

5. Thanks so much. It makes a lot of sense now.

So now, substituting e^y back in, I get

e^y = x / 1-5x

If I take the ln of both sides:

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

lne^y = y...therefore

y = ln (x/1-5x)