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If $\cosh ^{-1} x=2 \log _e(\sqrt{2}+1)$, then $x=$
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Verified Answer
The correct answer is:
3
We have, $\cosh ^{-1} x=2 \log _e(\sqrt{2}+1)$
$$
\begin{aligned}
& \Rightarrow \log _e\left(x+\sqrt{x^2-1}\right)=\log _e(\sqrt{2}+1)^2 \\
& \Rightarrow \quad x+\sqrt{x^2-1}=(\sqrt{2}+1)^2 \\
& \Rightarrow \quad x+\sqrt{x^2-1}=3+2 \sqrt{2}=3+\sqrt{8}
\end{aligned}
$$
On comparing rational and irrational part, we get
$$
x=3
$$
$$
\begin{aligned}
& \Rightarrow \log _e\left(x+\sqrt{x^2-1}\right)=\log _e(\sqrt{2}+1)^2 \\
& \Rightarrow \quad x+\sqrt{x^2-1}=(\sqrt{2}+1)^2 \\
& \Rightarrow \quad x+\sqrt{x^2-1}=3+2 \sqrt{2}=3+\sqrt{8}
\end{aligned}
$$
On comparing rational and irrational part, we get
$$
x=3
$$
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