Answer
$\frac{{13}}{2}{e^x} - \frac{3}{2}{e^{ - x}}$
Work Step by Step
$$\eqalign{
& 8\sinh x + 5\cosh x \cr
& {\text{Use the definition of hyperbolic functions }} \cr
& \sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}{\text{ and }}\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2} \cr
& {\text{Therefore}}{\text{,}} \cr
& 8\sinh x + 5\cosh x = 8\left( {\frac{{{e^x} - {e^{ - x}}}}{2}} \right) + 5\left( {\frac{{{e^x} + {e^{ - x}}}}{2}} \right) \cr
& {\text{Simplifying}} \cr
& = 4\left( {{e^x} - {e^{ - x}}} \right) + \frac{5}{2}\left( {{e^x} + {e^{ - x}}} \right) \cr
& = 4{e^x} - 4{e^{ - x}} + \frac{5}{2}{e^x} + \frac{5}{2}{e^{ - x}} \cr
& = \frac{{13}}{2}{e^x} - \frac{3}{2}{e^{ - x}} \cr} $$
