Express \(y=\sinh ^{-1}x\) as a natural logarithm.

Solution Since \(y=\sinh ^{-1}x\), where \(x=\sinh y,\) we have \[ \begin{eqnarray*} x &=&\dfrac{e^{y}-e^{-y}}{2} \\ 2xe^{y} &=&( e^{y}) ^{2}-1\qquad {\color{#0066A7}{\hbox{Multiply both sides by }2e^{y}.}} \\ ( e^{y}) ^{2}-2xe^{y}-1 &=&0 \end{eqnarray*} \]

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This is a quadratic equation in \(e^{y}\). Use the quadratic formula and solve for \(e^{y}\). \[ \begin{eqnarray*} e^{y} &=&\dfrac{2x\pm \ \sqrt{4x^{2}+4}}{2}\qquad \\ e^{y} &=&x\pm \ \sqrt{x^{2}+1}\qquad\ {\color{#0066A7}{\hbox{ Simplify.}}} \end{eqnarray*} \]

Since \(e^{y}>0\) and \(x<\ \sqrt{x^{2}+1}\) for all \(x\), the minus sign on the right side is not possible. As a result, \(e^{y}=x+\ \sqrt{x^{2}+1}\) so that \[ y=\sinh ^{-1}x=\ln ( x+\ \sqrt{x^{2}+1}) \]