Question 3.
Given the equation:
[tex]C(t)=9(0.5)^{0.021t}[/tex]
Where C is in milligrams per litre t minutes after taking the medicine.
Let's solve for the follosing:
• (a). Write down C(0).
To find C(0), substitute 0 for t and solve for c(0):
[tex]\begin{gathered} c(0)=9(0.5)^{0.021*0} \\ \\ c(0)=9(0.5)^0 \\ \\ c(0)=9(1) \\ \\ c(0)=9 \end{gathered}[/tex]
• (b). Find the concentration of the medication left in the patient's bloodstream after 40 minutes.
Substitute 40 for t and solve for C(40).
We have:
[tex]\begin{gathered} c(40)=9(0.5)^{0.021(40)} \\ \\ c(40)=9(0.5)^{0.84} \end{gathered}[/tex]
Solving further:
[tex]C(40)=5.02\text{ mg}[/tex]
Therefore, the concentration after 40 minutes is 5.03 milligrams per litre
• (c)., To solve this, first substitute 0.350 for C(t) and find t:
[tex]\begin{gathered} 0.350=9(0.5)^{0.021t} \\ \\ \end{gathered}[/tex]
Divide both sides by 9:
[tex]\begin{gathered} \frac{0.350}{9}=\frac{9(0.5)^{0.021t}}{9} \\ \\ \frac{0.350}{9}=0.5^{0.021t} \end{gathered}[/tex]
Take the natural loagraithm of both sides:
[tex]\begin{gathered} ln(\frac{350}{9})=0.021tln(0.5) \\ \\ −3.247046=0.021t(−0.693147) \\ \\ −3.247046=-0.014556t \\ \\ t=\frac{−3.247046}{-0.014556} \\ \\ t=223.07 \end{gathered}[/tex]
Therefore, the patient will take the medicine again 223 minutes after 14:00.
Convert the time to hours:
Where 60 mins = 1 hour
[tex]\frac{223}{60}=3\text{ hrs 43 mins}[/tex]
14:00 + 3:43 = 17:43
Therefore, the patient will take the medicine at 17:43
ANSWER:
(A). 9
(B). 5.02 mg per litre
(c). 17:43
