Answer
a) $\dfrac{13}{5}$;
b) $\dfrac{e^{10}+1}{2e^5}$
Work Step by Step
Apply definition of hyperbolic function for $\cosh x$
a) $\cosh ({\ln{5}})=\dfrac{e^{(\ln 5)}+e^{-(\ln 5)}}{2}$
or,$=\dfrac{5+\frac{1}{5}}{2}$
or, $=\dfrac{\dfrac{26}{5}}{2}$
or, $=\dfrac{13}{5}$
b) $\cosh 5=\dfrac{e^{(5)}+e^{-(5)}}{2}$
Thus,
$=\dfrac{e^5+\dfrac{1}{e^5}}{2}$
or, $=\dfrac{\dfrac{e^{10}+1}{e^5}}{2}$
$=\dfrac{e^{10}+1}{2e^5}$
