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If $\sin h(\log x)=-2$ then $x=$
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Verified Answer
The correct answer is:
$\sqrt{5}-2$
$\sin h(\log x)=-2$
Let $\log x=y \Rightarrow x=e^y$ $\sin h y=3$
$\begin{aligned}
& \Rightarrow \quad \frac{e^y-e^{-y}}{2}=-2 \\
& e^y-e^{\frac{1}{y}}=-4 \\
& \Rightarrow \quad\left(e^y\right)^2-1=-4 e^y \Rightarrow\left(e^y\right)^2+4 e^y-1=0 \\
& \Rightarrow \quad e^y=\frac{-4 \pm \sqrt{16+4}}{2}=\frac{-4 \pm 2 \sqrt{5}}{2} \\
& \Rightarrow \quad e^y=-2 \pm \sqrt{5} \\
& \because \quad e^y>0 \\
& \therefore \quad e^y=-2+\sqrt{5} \\
& \therefore \quad x=\sqrt{5}-2 .
\end{aligned}$
Let $\log x=y \Rightarrow x=e^y$ $\sin h y=3$
$\begin{aligned}
& \Rightarrow \quad \frac{e^y-e^{-y}}{2}=-2 \\
& e^y-e^{\frac{1}{y}}=-4 \\
& \Rightarrow \quad\left(e^y\right)^2-1=-4 e^y \Rightarrow\left(e^y\right)^2+4 e^y-1=0 \\
& \Rightarrow \quad e^y=\frac{-4 \pm \sqrt{16+4}}{2}=\frac{-4 \pm 2 \sqrt{5}}{2} \\
& \Rightarrow \quad e^y=-2 \pm \sqrt{5} \\
& \because \quad e^y>0 \\
& \therefore \quad e^y=-2+\sqrt{5} \\
& \therefore \quad x=\sqrt{5}-2 .
\end{aligned}$
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