If 2 cosh 2x + 10 sinh 2x = 5,then x = | vedclass

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(A) Given the equation: $2 \cosh 2x + 10 \sinh 2x = 5$. Using the definitions $\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}$ and $\sinh 2x = \frac{e^{2x} - e

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$\frac{1}{2} \log \frac{4}{3}$

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(A) Given the equation: $2 \cosh 2x + 10 \sinh 2x = 5$. Using the definitions $\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}$ and $\sinh 2x = \frac{e^{2x} - e^{-2x}}{2}$: $2 \left( \frac{e^{2x} + e^{-2x}}{2} \right) + 10 \left( \frac{e^{2x} - e^{-2x}}{2} \right) = 5$ $(e^{2x} + e^{-2x}) + 5(e^{2x} - e^{-2x}) = 5$ $6e^{2x} - 4e^{-2x} = 5$ Multiply by $e^{2x}$: $6(e^{2x})^2 - 5e^{2x} - 4 = 0$. Let $u = e^{2x}$,then $6u^2 - 5u - 4 = 0$. Factoring the quadratic: $(3u - 4)(2u + 1) = 0$. So,$u = \frac{4}{3}$ or $u = -\frac{1}{2}$. Since $e^{2x} > 0$,we have $e^{2x} = \frac{4}{3}$. Taking the natural logarithm on both sides: $2x = \ln \frac{4}{3}$. Therefore,$x = \frac{1}{2} \ln \frac{4}{3}$.

If $2 \cosh 2x + 10 \sinh 2x = 5$,then $x =$

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