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If $2 \sinh x=\cosh x$, then $x=$
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Verified Answer
The correct answer is:
$\frac{1}{2} \log 3$
Given $2 \sinh x=\cos h x$
$\begin{aligned}
& \Rightarrow 2\left(\frac{\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}}{2}\right)=\frac{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}{2} \Rightarrow 2 \mathrm{e}^{\mathrm{x}}-2 \mathrm{e}^{-\mathrm{x}}=\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}} \\
& \Rightarrow \mathrm{e}^{\mathrm{x}}=3 \mathrm{e}^{-\mathrm{x}} \Rightarrow \mathrm{e}^{2 \mathrm{x}}=3 \Rightarrow \mathrm{x}=\frac{1}{2} \log _{\mathrm{e}} 3
\end{aligned}$
$\begin{aligned}
& \Rightarrow 2\left(\frac{\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}}{2}\right)=\frac{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}{2} \Rightarrow 2 \mathrm{e}^{\mathrm{x}}-2 \mathrm{e}^{-\mathrm{x}}=\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}} \\
& \Rightarrow \mathrm{e}^{\mathrm{x}}=3 \mathrm{e}^{-\mathrm{x}} \Rightarrow \mathrm{e}^{2 \mathrm{x}}=3 \Rightarrow \mathrm{x}=\frac{1}{2} \log _{\mathrm{e}} 3
\end{aligned}$
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