Answer
a) $k \approx-0.000120968$
b) $ t \approx24765$
Work Step by Step
a) We have: $x=2xe^{kt}$
Plug in $t=1690$ to obtain:
$x=2xe^{k\cdot1690}\\0.5=e^{5730k}\\5730k=\ln0.5\\ k=\dfrac{\ln0.5}{5730}\approx-0.000120968$
b) We have: $x=2xe^{kt}$
In order to find the value of $t$, we need to plug into the above equation $k=0.05$ to obtain:
$0.05=e^{-0.000120968(t)}\\ -0.000120968t=\ln0.05\\t=\dfrac{\ln0.05}{-0.000120968}\approx24765$
