Answer
about 5.5 hours
Work Step by Step
The decay model: $A=A_{0}e^{kt}$ ($ k < 0 )$
We find k from the half life information
$0.5A_{0}=A_{0}e^{k\cdot 36}\qquad/\div A_{0}$
$0.5=e^{36k}\qquad $ ... apply ln( ) to both sides
$\ln 0.5=36k$
$ k=\displaystyle \frac{\ln 0.5}{36}\approx$-0.0192540883489$\approx-0.0193$
The decay model is : $A=A_{0}e^{-0.0193t}$
We want t (in hours) when $A=0.90A_{0}$
$0.9A_{0}=A_{0}e^{-0.0193t}\qquad/\div A_{0}$
$0.9=e^{-0.0193t}\qquad $ ... apply ln( ) to both sides
$\ln 0.9=-0.0193t\qquad \div(-0.0193)$
$ t=\displaystyle \frac{\ln 0.9}{-0.0193}\approx$5.459094075548$\approx 5.5$ hours