Answer
a) $$\frac{13}{5}$$
b) $$\frac{e^{10}+1}{2e^5}$$
Work Step by Step
Remember that:
$$\cosh x=\frac{e^x+e^{-x}}{2}$$
a) $\cosh ({\ln{5}})$
$=\frac{e^{(\ln 5)}+e^{-(\ln 5)}}{2}
=\frac{5+\frac{1}{5}}{2}
=\frac{\frac{26}{5}}{2}
=\frac{26}{10}
=\frac{13}{5}$
b) $\cosh 5$
$=\frac{e^{(5)}+e^{-(5)}}{2}
=\frac{e^5+\frac{1}{e^5}}{2}
=\frac{\frac{e^{10}+1}{e^5}}{2}
=\frac{e^{10}+1}{2e^5}$
