Find the inverse of
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
$\displaystyle \displaystyle \begin{align*} x &= \frac{2^y}{1 + 2^y} \\ x&= \frac{1 + 2^y - 1}{1 + 2^y} \\ x &= \frac{1 + 2^y}{1 + 2^y } - \frac{1}{1 + 2^y} \\ x &= 1 - \frac{1}{1 + 2^y} \\ \frac{1}{1 + 2^y} &= 1 - x \\ 1 + 2^y &= \frac{1}{1 - x} \\ 2^y &= \frac{1}{1 - x} - 1 \\ 2^y &= \frac{1}{1 - x} - \frac{1 - x}{1 - x} \\ 2^y &= \frac{1 - (1 - x)}{1 - x} \\ 2^y &= \frac{x}{1 - x} \\ \ln{\left(2^y\right)} &= \ln{\left(\frac{x}{1 - x}\right)} \\ y\ln{2} &= \ln{x} - \ln{(1 - x)} \\ y &= \frac{\ln{x} - \ln{(1 - x)}}{\ln{2}} \end{align*}$
Hello, Remriel!
$\displaystyle \text{Find the inverse function: }\:f(x) \:=\:\frac{2^x}{1+2^x}$
I know the answer . . . . I'm always impressed by that!
We are given: .$\displaystyle y \;=\;\frac{2^x}{1 + 2^x}$
Switch variables: .$\displaystyle x \;=\;\frac{2^y}{1+2^y}$
$\displaystyle \text{Solve for }y\!:\;\;x(1+2^y) \;=\;2^y$
m . . . . . . . . $\displaystyle x + x\!\cdot\!2^y \;=\;2^y$
. . . . . . . . . $\displaystyle 2^y - x\!\cdot\!2^y \;=\;x$
. . . . . . . . . $\displaystyle 2^y(1-x) \;=\;x$
. . . . . . . . . . . . . . $\displaystyle 2^y \;=\;\frac{x}{1-x}$
Take logs, base 2: .$\displaystyle \log_2(2^y) \;=\;\log_2\!\left(\frac{x}{1-x}\right) $
. . . . . . . . . . . . . $\displaystyle y\!\cdot\!\underbrace{\log_2(2)}_{\text{This is 1}} \;=\;\log_2\!\left(\frac{x}{1-x}\right)$
Therefore: .$\displaystyle y \;=\;\log_2(x) - \log_2(1-x)$
We can manipulate Prove It's answer to get your one using the log laws.
By the subtraction rule: $\displaystyle y = \dfrac{\ln(x) - \ln(1-x)}{\ln(2)} = \dfrac{\ln \left(\frac{x}{1-x}\right)}{\ln(2)}$
By the change of base rule: $\displaystyle \log_2(y) = \dfrac{\ln(y)}{\ln(2)}$ of which the RHS is the safe as the expression above.
Putting in the expression for y: $\displaystyle \log_2 \left(\dfrac{x}{1-x}\right)$
Thus the two answers are equivalent.
Also note that for the inverse $\displaystyle x \neq 0, 1$ - what does that tell you about the original function's range (admittedly you don't need to include this bit for your question)