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The inverse of the function $y=\frac{10^x-10^{-x}}{10^x+10^{-x}}$ is
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Verified Answer
The correct answer is:
$\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right)$
We have,
$$
\begin{array}{rlrl}
y & & =\frac{10^x-10^{-x}}{10^x+10^{-x}} \\
\Rightarrow \quad y & =\frac{10^{2 x}-1}{10^{2 x}+1} \\
\Rightarrow \quad 10^{2 x} y+y & =10^{2 x}-1 \\
\Rightarrow \quad 10^{2 x}(y-1) & =-(y+1) \\
\Rightarrow \quad & 10^{2 x} & =\frac{y+1}{1-y} \\
\Rightarrow \quad & 2 x & =\log _{10} \frac{y+1}{1-y} \\
\Rightarrow \quad x & =\frac{1}{2} \log _{10}\left(\frac{1+y}{1-y}\right) \\
\therefore \quad & f^{-1}(x) & =\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right)
\end{array}
$$
$$
\begin{array}{rlrl}
y & & =\frac{10^x-10^{-x}}{10^x+10^{-x}} \\
\Rightarrow \quad y & =\frac{10^{2 x}-1}{10^{2 x}+1} \\
\Rightarrow \quad 10^{2 x} y+y & =10^{2 x}-1 \\
\Rightarrow \quad 10^{2 x}(y-1) & =-(y+1) \\
\Rightarrow \quad & 10^{2 x} & =\frac{y+1}{1-y} \\
\Rightarrow \quad & 2 x & =\log _{10} \frac{y+1}{1-y} \\
\Rightarrow \quad x & =\frac{1}{2} \log _{10}\left(\frac{1+y}{1-y}\right) \\
\therefore \quad & f^{-1}(x) & =\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right)
\end{array}
$$
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