The inverse of the function $y=\frac{10^x-10^{-x}}{10^x+10^{-x}}$ is

The inverse of the function $y=\frac{10^x-10^{-x}}{10^x+10^{-x}}$ is
  1. $\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right)$
  2. $\frac{1}{2} \log _{10}\left(\frac{2+x}{2-x}\right)$
  3. $\frac{1}{2} \log _{10}\left(\frac{1-x}{1+x}\right)$
  4. $\frac{1}{2} \log _{10}\left(\frac{2-x}{2+x}\right)$

Solution

We have, $ \begin{array}{rlrl} y & & =\frac{10^x-10^{-x}}{10^x+10^{-x}} \\ \Rightarrow \quad y & =\frac{10^{2 x}-1}{10^{2 x}+1} \\ \Rightarrow \quad 10^{2 x} y+y & =10^{2 x}-1 \\ \Rightarrow \quad 10^{2 x}(y-1) & =-(y+1) \\ \Rightarrow \quad & 10^{2 x} & =\frac{y+1}{1-y} \\ \Rightarrow \quad & 2 x & =\log _{10} \frac{y+1}{1-y} \\ \Rightarrow \quad x & =\frac{1}{2} \log _{10}\left(\frac{1+y}{1-y}\right) \\ \therefore \quad & f^{-1}(x) & =\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right) \end{array} $

Asked in: AP EAMCET 2021 (25 Aug Shift 1)

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