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))) \\ \\ (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=(\ln0.7)/(-0.35) \\ \\ -t=(\ln0.7)/(-0.35)-12 \\ \\ t=12-(\ln0.7)/(-0.35) \\ \\ t\cong10.98 \\ \\ t\cong11 \end{gathered}`

The percent will reach 30% in 11 hours.