Answer
a) $\approx 0.262$
b) $\approx 3.816$ years
$c) \approx 11.431$ years
Work Step by Step
a) The exponential growth can be written as: $A=A_0e^{-kt}$ ...(1)
This implies that $\dfrac{1}{2} A_0=A_0e^{-2.645k} \implies k =-\dfrac{\ln (2)}{2.645}$
or, $k\approx 0.262$
b) From part (a), we have $k\approx 0.262$ years and $\dfrac{1}{k} \approx 3.816$ years
c) The exponential growth can be written as: $A=A_0e^{-kt}$ ...(1)
This implies that $(0.05)A=Ae^{-\frac{\ln (2)}{2.645}t} \implies t =\dfrac{2.645\ln (20)}{\ln 2}$
or, $t\approx 11.431$ years
Hence, a) $\approx 0.262$ b) $\approx 3.816$ years c) $\approx 11.431$ years
