If sinh(log x) = -2,then x = | vedclass

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(A) Given $\sinh(\log x) = -2$. Let $\log x = y$,then $x = e^y$. Using the definition $\sinh y = \frac{e^y - e^{-y}}{2} = -2$. $e^y - e^{-y} = -4$. Mu

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$\sqrt{5} - 2$

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(A) Given $\sinh(\log x) = -2$. Let $\log x = y$,then $x = e^y$. Using the definition $\sinh y = \frac{e^y - e^{-y}}{2} = -2$. $e^y - e^{-y} = -4$. Multiply by $e^y$: $(e^y)^2 - 1 = -4e^y$. $(e^y)^2 + 4e^y - 1 = 0$. Using the quadratic formula $e^y = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{-4 \pm \sqrt{20}}{2} = -2 \pm \sqrt{5}$. Since $x = e^y$ must be positive,$e^y = \sqrt{5} - 2$. Thus,$x = \sqrt{5} - 2$.

If $\sinh(\log x) = -2$,then $x =$

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