Answer
a. $k=0.036$
b. $A=60.39$,
c. $2019$
Work Step by Step
$A=35.3e^{kt}$,
where, $t=0, A=35.3$
whereas $t=10, A=50.5$,
a. $35.3e^{10k}=50.5 $,
$e^{10k}=1.4305949$,
$10k= \ln (1.4305949)$
$k=0.0358090$,
$k=0.0358$, or $k=0.036$
b. $A=35.3e^{15k}=60.39$
c. $70= 35.3e^{0.036t}$,
$1.98=e^{0.036t}$,
$\ln (1.98)= 0.036t$,
$t=18.97$,
$t=19^{th} year$
The year $2000+19=2019$.
