Question

The inverse of the function $y=\frac{2^x}{1+2^x}$ us :

• A. $x={\log }_2\frac{y}{1-y}$
• B. $x={\log }_2\frac{y}{1-2^y}$
• C. $x={\log }_2(1-\frac{1}{y})$
• D. $x={\log }_2(\frac{1}{1-y})$

Answer 1

Answer: (A)

Given,
$y=\frac{2^x}{1+2^x}$
$\Rightarrow y+2^{x}y=2^x$
$\Rightarrow y=2^x(1-y)$
$\Rightarrow 2^x=\frac{y}{1-y}$
Taking log both side at base 2, we get
${\log }_2\left(2^x\right)={\log }_2\frac{y}{1-y}$
$\Rightarrow x={\log }_2\frac{y}{1-y}$
$\therefore$ Inverse $={\log }_2\frac{y}{1-y}$