essary: \( 3 \cosh x+5 \sinh x=8 \)

To solve the equation \( 3 \cosh x + 5 \sinh x = 8 \), we can use the definitions of hyperbolic functions: \[ \cosh x = \frac{e^x + e^{-x}}{2}, \quad \sinh x = \frac{e^x - e^{-x}}{2} \] Substituting these definitions into the equation gives: \[ 3 \left( \frac{e^x + e^{-x}}{2} \right) + 5 \left( \frac{e^x - e^{-x}}{2} \right) = 8 \] Multiplying through by 2 to eliminate the fractions: \[ 3(e^x + e^{-x}) + 5(e^x - e^{-x}) = 16 \] Distributing the terms: \[ 3e^x + 3e^{-x} + 5e^x - 5e^{-x} = 16 \] Combining like terms: \[ (3e^x + 5e^x) + (3e^{-x} - 5e^{-x}) = 16 \] This simplifies to: \[ 8e^x - 2e^{-x} = 16 \] To eliminate \( e^{-x} \), we can multiply the entire equation by \( e^x \): \[ 8e^{2x} - 2 = 16e^x \] Rearranging gives us a standard quadratic form: \[ 8e^{2x} - 16e^x - 2 = 0 \] Letting \( y = e^x \), we rewrite the equation as: \[ 8y^2 - 16y - 2 = 0 \] Now we can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 8 \), \( b = -16 \), and \( c = -2 \): \[ b^2 - 4ac = (-16)^2 - 4 \cdot 8 \cdot (-2) = 256 + 64 = 320 \] Now substituting back into the quadratic formula: \[ y = \frac{16 \pm \sqrt{320}}{16} \] Calculating \( \sqrt{320} \): \[ \sqrt{320} = \sqrt{64 \cdot 5} = 8\sqrt{5} \] Thus, we have: \[ y = \frac{16 \pm 8\sqrt{5}}{16} = 1 \pm \frac{\sqrt{5}}{2} \] This gives us two possible values for \( y \): \[ y_1 = 1 + \frac{\sqrt{5}}{2}, \quad y_2 = 1 - \frac{\sqrt{5}}{2} \] Since \( y = e^x \) must be positive, we discard \( y_2 \) if it is negative. We need to check if \( 1 - \frac{\sqrt{5}}{2} \) is positive: Calculating \( \frac{\sqrt{5}}{2} \approx 1.118 \), we find: \[ 1 - \frac{\sqrt{5}}{2} \approx 1 - 1.118 = -0.118 \quad (\text{not valid}) \] Thus, we only consider: \[ y = 1 + \frac{\sqrt{5}}{2} \] Now, we can find \( x \): \[ e^x = 1 + \frac{\sqrt{5}}{2} \] Taking the natural logarithm: \[ x = \ln\left(1 + \frac{\sqrt{5}}{2}\right) \] Thus, the solution to the equation \( 3 \cosh x + 5 \sinh x = 8 \) is: \[ \boxed{\ln\left(1 + \frac{\sqrt{5}}{2}\right)} \]