Answer
a) $\approx 0.262$ b) $\approx 3.816$ years c) $\approx 11.431$ years
Work Step by Step
a) Consider the exponential growth equation as follows: $A=A_0e^{-kt}$
Then, we have $(\dfrac{1}{2})A_0=A_0e^{-(2.645)k}$
or, $k =-\dfrac{\ln (2)}{2.645} \approx 0.262$
b) Take the help of part (a), we get
$k\approx 0.262$ years and $\dfrac{1}{k} \approx 3.816$ years
c) Consider the exponential growth equation as follows: $A=A_0e^{-kt}$
This implies that $(0.05)A=Ae^{-[\frac{\ln (2)}{2.645}]t}$
or, $t =\dfrac{2.645\ln (20)}{\ln (2)}$
Thus, $t\approx 11.431$ years
