Calculating the inverse of $2^x/(2^x+1)$

Question

I need help calculating the inverse of $$f(x)=\frac{2^x}{2^x+1}$$

I know this should be done using logarithms.

I started with $\log_2(x)=\log_2(\frac{2^y}{2^y+1})$ but don't know how to proceed.

Any tips on how to continue?

Thank you!

Answer 1

$$y = \frac{2^x}{2^x+1}$$

Multiply for the denominator both sides:

$$y2^x + y = 2^x$$

Collect the $x$ terms:

$$2^x(y-1) = -y$$

Hence

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

Taking the log base $2$ ($\lg$) we get

$$\lg 2^x = \lg\left(-\frac{y}{y-1}\right) = \lg(-y) - \lg(y-1)$$

Finally

$$x = \lg(-y) - \lg(y-1)$$

Answer 2

hint: $y = \dfrac{2^x}{1+2^x}$, switch $x$ and $y$ in the equation,and solve for $y$. Can you continue?

Answer 3

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

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

$ x = \log_2 (\frac{1}{f(x)-1}-1)$