Given that $y = sin h^{- 1} x$ , find a quadratic equation satisfied by $e^{y}$ . Hence obtain the logarithmic form of $sin h^{- 1} x$ . Explain why you discard one of the solution.

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Solution

Given $y = s i n h^{- 1} x \Rightarrow x = s i n h y$

$\Rightarrow x = \frac{e^{y} - e^{- y}}{2}$

$\Rightarrow x = \frac{e^{2 y} - 1}{2 e^{y}}$

$\Rightarrow (e^{y})^{2} - e^{y} (2 x) - 1 = 0$ $\Rightarrow$ quadratic equation satisfied by $e^{y}$

$\Rightarrow e^{y} = x \pm \sqrt{x^{2} + 1}$

As $e^{y}$ is always positive, we will take only plus sign

$\Rightarrow y = l n (x + \sqrt{x^{2} - 1})$

$\Rightarrow y = s i n h^{- 1} x = l n (x + \sqrt{x^{2} + 1})$

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