If 4+6(e^{2x}+1) tanh x = 11 cosh x + 11 sinh | vedclass

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(D) Given equation: $4+6(e^{2x}+1) 	anh x = 11 \cosh x + 11 \sinh x$Substitute $	anh x = \frac{e^x-e^{-x}}{e^x+e^{-x}}$,$\cosh x = \frac{e^x+e^{-x}}

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$\log_e 2$

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(D) Given equation: $4+6(e^{2x}+1) 	anh x = 11 \cosh x + 11 \sinh x$Substitute $	anh x = \frac{e^x-e^{-x}}{e^x+e^{-x}}$,$\cosh x = \frac{e^x+e^{-x}}{2}$,and $\sinh x = \frac{e^x-e^{-x}}{2}$:$4+6(e^{2x}+1) \left(\frac{e^x-e^{-x}}{e^x+e^{-x}}ight) = 11 \left(\frac{e^x+e^{-x}}{2}ight) + 11 \left(\frac{e^x-e^{-x}}{2}ight)$$4+6(e^{2x}+1) \left(\frac{e^{2x}-1}{e^{2x}+1}ight) = 11 \left(\frac{2e^x}{2}ight)$$4+6(e^{2x}-1) = 11e^x$$6e^{2x} - 11e^x - 2 = 0$Let $e^x = t$,where $t > 0$:$6t^2 - 11t - 2 = 0$$(6t+1)(t-2) = 0$$t = 2$ or $t = -\frac{1}{6}$Since $t > 0$,we have $e^x = 2$,which implies $x = \log_e 2$.

If $4+6(e^{2x}+1) \tanh x = 11 \cosh x + 11 \sinh x$,then $x=$

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