# Thread: Inverse of Exponential Function

1. ## Inverse of Exponential Function

Find the inverse of

$f(x)=\frac{2^x}{1+2^x}$

I know the answer, I'm just not sure how to get to it.

2. ## Re: Inverse of Exponential Function

Originally Posted by Remriel
$f(x)=\frac{2^x}{1+2^x}$

I know the answer, I'm just not sure how to get to it.
For the inverse function, the x and y values swap, so you'll get

\displaystyle \begin{align*} x &= \frac{2^y}{1 + 2^y} \\ x&= \frac{1 + 2^y - 1}{1 + 2^y} \\ x &= \frac{1 + 2^y}{1 + 2^y } - \frac{1}{1 + 2^y} \\ x &= 1 - \frac{1}{1 + 2^y} \\ \frac{1}{1 + 2^y} &= 1 - x \\ 1 + 2^y &= \frac{1}{1 - x} \\ 2^y &= \frac{1}{1 - x} - 1 \\ 2^y &= \frac{1}{1 - x} - \frac{1 - x}{1 - x} \\ 2^y &= \frac{1 - (1 - x)}{1 - x} \\ 2^y &= \frac{x}{1 - x} \\ \ln{\left(2^y\right)} &= \ln{\left(\frac{x}{1 - x}\right)} \\ y\ln{2} &= \ln{x} - \ln{(1 - x)} \\ y &= \frac{\ln{x} - \ln{(1 - x)}}{\ln{2}} \end{align*}

3. ## Re: Inverse of Exponential Function

How did you get -1 in the numerator in the second equation?

Also, your answer is different from the book which says: $y=\log_{2}(\frac{x}{1-x})$.

4. ## Re: Inverse of Exponential Function

Originally Posted by Remriel
How did you get -1 in the numerator in the second equation?
Added a cleverly disguised 0, in this case, 1 - 1...

5. ## Re: Inverse of Exponential Function

Originally Posted by Remriel

Also, your answer is different from the book which says: $y=\log_{2}(\frac{x}{1-x})$.
It doesn't matter which base of the logarithm you use (just choose the most convenient).
So try the same thing like Prove it did but take 2 as the base of the logarithm.

6. ## Re: Inverse of Exponential Function

Hello, Remriel!

$\text{Find the inverse function: }\:f(x) \:=\:\frac{2^x}{1+2^x}$

I know the answer . . . .
I'm always impressed by that!

We are given: . $y \;=\;\frac{2^x}{1 + 2^x}$

Switch variables: . $x \;=\;\frac{2^y}{1+2^y}$

$\text{Solve for }y\!:\;\;x(1+2^y) \;=\;2^y$

m . . . . . . . . $x + x\!\cdot\!2^y \;=\;2^y$

. . . . . . . . . $2^y - x\!\cdot\!2^y \;=\;x$

. . . . . . . . . $2^y(1-x) \;=\;x$

. . . . . . . . . . . . . . $2^y \;=\;\frac{x}{1-x}$

Take logs, base 2: . $\log_2(2^y) \;=\;\log_2\!\left(\frac{x}{1-x}\right)$

. . . . . . . . . . . . . $y\!\cdot\!\underbrace{\log_2(2)}_{\text{This is 1}} \;=\;\log_2\!\left(\frac{x}{1-x}\right)$

Therefore: . $y \;=\;\log_2(x) - \log_2(1-x)$

7. ## Re: Inverse of Exponential Function

Originally Posted by Remriel
How did you get -1 in the numerator in the second equation?

Also, your answer is different from the book which says: $y=\log_{2}(\frac{x}{1-x})$.
We can manipulate Prove It's answer to get your one using the log laws.

By the subtraction rule: $y = \dfrac{\ln(x) - \ln(1-x)}{\ln(2)} = \dfrac{\ln \left(\frac{x}{1-x}\right)}{\ln(2)}$

By the change of base rule: $\log_2(y) = \dfrac{\ln(y)}{\ln(2)}$ of which the RHS is the safe as the expression above.

Putting in the expression for y: $\log_2 \left(\dfrac{x}{1-x}\right)$

Thus the two answers are equivalent.

Also note that for the inverse $x \neq 0, 1$ - what does that tell you about the original function's range (admittedly you don't need to include this bit for your question)

8. ## Re: Inverse of Exponential Function

I understand now. Thanks everyone.