The solution of the equation 2 cosh 2x + 10 s | vedclass

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

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$\frac{1}{2} \log \left(\frac{4}{3}\right)$

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(B) Given the equation: $2 \cosh 2x + 10 \sinh 2x = 5$ Substitute the exponential definitions $\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}$ and $\sinh 2x = \frac{e^{2x} - e^{-2x}}{2}$: $2 \left(\frac{e^{2x} + e^{-2x}}{2}\right) + 10 \left(\frac{e^{2x} - e^{-2x}}{2}\right) = 5$ $(e^{2x} + e^{-2x}) + 5(e^{2x} - e^{-2x}) = 5$ $e^{2x} + e^{-2x} + 5e^{2x} - 5e^{-2x} = 5$ $6e^{2x} - 4e^{-2x} = 5$ Multiply by $e^{2x}$: $6(e^{2x})^2 - 5e^{2x} - 4 = 0$ Let $u = e^{2x}$,then $6u^2 - 5u - 4 = 0$ Factor the quadratic: $6u^2 - 8u + 3u - 4 = 0 \Rightarrow 2u(3u - 4) + 1(3u - 4) = 0$ $(2u + 1)(3u - 4) = 0$ Since $u = e^{2x} > 0$,we must have $3u - 4 = 0$,so $u = \frac{4}{3}$ $e^{2x} = \frac{4}{3} \Rightarrow 2x = \log \left(\frac{4}{3}\right)$ $x = \frac{1}{2} \log \left(\frac{4}{3}\right)$

The solution of the equation $2 \cosh 2x + 10 \sinh 2x = 5$ is

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