Question

If a substance is injected into the bloodstream, the percent of the maximum dosage that is present at time t is given by y=100{1−e^[−0.35(12−t)]}, where t is in hours, with 0≤t≤12. In how many hours will the percent reach 30% ?

Answer 1

In order to find the time t when the percent y will reach 30%, we need to replace y by 30, and solve the equation for t:

$\begin{gathered} y=100(1-e^{-0.35(12-t)}) \\ \\ 30=100(1-e^{-0.35(12-t)}) \\ \\ \frac{30}{100}=1-e^{-0.35(12-t)} \\ \\ 0.3+e^{-0.35(12-t)}=1 \\ \\ e^{-0.35(12-t)}=1-0.3 \\ \\ e^{-0.35(12-t)}=0.7 \\ \\ \ln e^{-0.35(12-t)}=\ln 0.7 \\ \\ -0.35(12-t)=\ln 0.7 \\ \\ 12-t=\frac{\ln0.7}{-0.35} \\ \\ -t=\frac{\ln0.7}{-0.35}-12 \\ \\ t=12-\frac{\ln0.7}{-0.35} \\ \\ t\cong10.98 \\ \\ t\cong11 \end{gathered}$

The percent will reach 30% in 11 hours.