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The solution of the equation $2 \cosh 2 x+10 \sinh 2 x=5$ is
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Verified Answer
The correct answer is:
$\frac{1}{2} \log \left(\frac{4}{3}\right)$
we have,
$$
\begin{gathered}
2 \cosh 2 x+10 \sinh 2 x=5 \\
2\left(\frac{e^{2 x}+e^{-2 x}}{2}\right)+10\left(\frac{e^{2 x}-e^{-2 x}}{2}\right)=5 \\
2 e^{2 x}+2 e^{-2 x}+10 e^{2 x}-10 e^{-2 x}=10 \\
12 e^{2 x}-8 e^{-2 x}-10=0 \\
6 e^{2 x}-4^{-2 x}-5=0 \\
6\left(e^{2 x}\right)^2-5 e^{2 x}-4=0 \\
6\left(e^{2 x}\right)^2-8 e^{2 x}+3 e^{2 x}-4=0 \\
\Rightarrow \quad 3 e^{2 x}-4=0,2 e^{2 x}+1 \neq 0 \\
\Rightarrow \quad e^{2 x}=\frac{4}{3} \\
\Rightarrow \quad 2 x=\log \frac{4}{3} \\
\Rightarrow \quad x=\frac{1}{2} \log \frac{4}{3}
\end{gathered}
$$
$$
\begin{gathered}
2 \cosh 2 x+10 \sinh 2 x=5 \\
2\left(\frac{e^{2 x}+e^{-2 x}}{2}\right)+10\left(\frac{e^{2 x}-e^{-2 x}}{2}\right)=5 \\
2 e^{2 x}+2 e^{-2 x}+10 e^{2 x}-10 e^{-2 x}=10 \\
12 e^{2 x}-8 e^{-2 x}-10=0 \\
6 e^{2 x}-4^{-2 x}-5=0 \\
6\left(e^{2 x}\right)^2-5 e^{2 x}-4=0 \\
6\left(e^{2 x}\right)^2-8 e^{2 x}+3 e^{2 x}-4=0 \\
\Rightarrow \quad 3 e^{2 x}-4=0,2 e^{2 x}+1 \neq 0 \\
\Rightarrow \quad e^{2 x}=\frac{4}{3} \\
\Rightarrow \quad 2 x=\log \frac{4}{3} \\
\Rightarrow \quad x=\frac{1}{2} \log \frac{4}{3}
\end{gathered}
$$
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