Many prescribed drugs must reach a “maintenance level” in the bloodstream to be effective. Say a person takes 300 milligrams of wonder drug Excellente´ per day and that whatever level of Excellente´ is in the bloodstream, p% is eliminated in one 24-hour period.

Find the approximate level of drugs during the first week.

The amount of drug in milligrams a person takes D=300 mg.

The percentage p of the drug eliminated from the blood stream after 24 hours = 60

$$\begin{gathered} \mathrm{The}\mathrm{daily}\mathrm{dose}\mathrm{of}\mathrm{the}\mathrm{medicine}\mathrm{is}\mathrm{D}. \\ \mathrm{Let}{\mathrm{L}}_1\mathrm{is}\mathrm{the}\mathrm{amount}\mathrm{of}\mathrm{drug}\mathrm{in}\mathrm{the}\mathrm{blood}\mathrm{stream}\mathrm{at}\mathrm{the}\mathrm{start}. \\ \mathrm{Hence},{\mathrm{L}}_1=\mathrm{D} \\ \mathrm{After}24\mathrm{hours}50\mathrm{\%}\mathrm{of}\mathrm{the}\mathrm{drug}\mathrm{has}\mathrm{been}\mathrm{eliminated}\mathrm{from}\mathrm{the}\mathrm{blood}\mathrm{stream}\mathrm{and}\mathrm{the}\mathrm{second}\mathrm{dose}\mathrm{is}\mathrm{taken}. \\ Hence,L2=(1-\frac{p}{100})L_1+D......\left(1\right) \\ \mathrm{In}\mathrm{general},\mathrm{after}kdays.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting}k=1in\left(2\right) \\ \mathrm{Therefore},L_2=(1-\frac{60}{100})\left(300\right)+300 \\ Simplify, \\ {L}_2=(1-0.60)\left(300\right)+300 \\ {L}_2=(0.40)\left(300\right)+300 \\ =40+100 \\ {L}_2=140 \end{gathered}$$

$$\begin{gathered} L_3=(1-\frac{p}{100})L_1+D......\left(1\right) \\ \mathrm{In}\mathrm{general},\mathrm{after}kdays.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting}k=2in\left(2\right) \\ \mathrm{Therefore},L_2=140 \\ Simplify, \\ {L}_3=(1-0.60)\left(140\right)+300 \\ {L}_3=(0.40)\left(140\right)+300 \\ {L}_3=356 \end{gathered}$$

$$\begin{gathered} L_4=(1-\frac{p}{100})L_3+D......\left(1\right) \\ \mathrm{In}\mathrm{general},\mathrm{after}kdays.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting}k=3in\left(2\right) \\ \mathrm{Therefore},L_3=356 \\ Simplify, \\ {L}_4=(1-0.60)\left(356\right)+300 \\ {L}_4=(0.40)\left(356\right)+300 \\ {L}_4=442.40 \end{gathered}$$

$$\begin{gathered} L_5=(1-\frac{p}{100})L_4+D......\left(1\right) \\ \mathrm{In}\mathrm{general},\mathrm{after}kdays.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting}k=4in\left(2\right) \\ \mathrm{Therefore},L_4=442.40 \\ Simplify, \\ {L}_5=(1-0.60)(442.40)+300 \\ {L}_5=(0.40)(442.40)+300 \\ {L}_5=479.96 \end{gathered}$$

$$\begin{gathered} L_6=(1-\frac{p}{100})L_5+D......\left(1\right) \\ \mathrm{In}\mathrm{general},\mathrm{after}kdays.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting}k=5in\left(2\right) \\ \mathrm{Therefore},L_5=476.96 \\ \mathrm{Simplify}, \\ {L}_6=(1-0.60)(476.96)+300 \\ {L}_6=(0.40)(476.96)+300 \\ {L}_6=490.784 \end{gathered}$$

$$\begin{gathered} L_7=(1-\frac{p}{100})L_6+D......\left(1\right) \\ \mathrm{In}\mathrm{general},\mathrm{after}kdays.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting}k=5in\left(2\right) \\ \mathrm{Therefore},L_6=490.78 \\ \mathrm{Simplify}, \\ {L}_7=(1-0.60)(490.78)+300 \\ {L}_7=(0.40)(490.78)+300 \\ {L}_7=496.31 \end{gathered}$$