10 December 2021

Statement

Express \(x\) as a function of \(y\).

\(y=\frac{1-e^{-x}}{1+e^{-x}}\)

Solution

\(y=\frac{1-e^{-x}}{1+e^{-x}}\)

\(\Leftrightarrow y(1+e^{-x})=1-e^{-x} \)

\(\Leftrightarrow y + ye^{-x}=1-e^{-x} \)

\(\Leftrightarrow 1 - y = ye^{-x} + e^{-x} \)

\(\Leftrightarrow 1 - y = e^{-x}(1 + y) \)

It might not seem obvious but the codomain of the original expression \(y=\frac{1-e^{-x}}{1+e^{-x}}\) is actually \([-1, 1]\) which means the LHS of the expression \(1 - y\) varies between \([0, 2]\) so it is safe to use the \(log\) function to remove the \(x\) from the exponent on the RHS.

\(\Leftrightarrow \ln{(1 - y)} = \ln{(e^{-x}(1 + y))} \)

\(\Leftrightarrow \ln{(1 - y)} = \ln{(1 + y)} + \ln{e^{-x}} \)

\(\Leftrightarrow \ln{(1 - y)} = \ln{(1 + y)} - x \)

\(\Leftrightarrow x = \ln{(1 + y)} - \ln{(1 - y)} \)