Given the function:
[tex]A(t)=40(0.83)^t[/tex]Where A(t) shows the amount of drug in a body after t hours.
Let's solve for the following:
• (a). Initial dosage:
Apply the exponential functions:
[tex]f(x)=a(b)^x[/tex]Where:
a is the initial value
b is the change factor.
Thus, we have the following:
a = 40
b = 0.83
Therefore, the initial dose is 40 mg.
• (b). What percent leaves the body each hour?
Apply the function:
[tex]f(x)=a(1-r)^x[/tex]Where:
r is the decay rate.
Thus, we have:
b = 1 - r
r = 1 - b
r = 1 - 0.83
r = 0.17
The percent that leaves the body each hour will be:
0.17 x 100 = 17%
Therefore, 17 percent of the drug leaves the body each hour.
• (c). What amount of drug is left after 12 hours?
Substitute 12 for t and solve for A(12):
[tex]\begin{gathered} A(12)=40(0.83)^{12} \\ \\ A(12)=40(0.1068900077) \\ \\ A(12)=4.28 \end{gathered}[/tex]The amount left after 12 hours is 4.28 mg.
• (d). The first whole number of hours at which there is less than 6 mg left.
Plug in 5.9 for A(t) and solve for t.
[tex]5.9=40(0.83)^t[/tex]Divide both sides by 40:
[tex]\begin{gathered} \frac{5.9}{40}=\frac{40(0.83)^t}{40} \\ \\ 0.1475=(0.83)^t \end{gathered}[/tex]Take the natural logarithm of both sides:
[tex]\begin{gathered} ln(0.1475)=tln(0.83) \\ \\ t=\frac{ln(0.1475)}{ln(0.83)} \\ \\ t=10.2 \end{gathered}[/tex]Therefore, the first whole number of hours where there is less than 6 mg left is 10 hours.
ANSWER:
• (a) 40 mg
,• (b) 17%
,• (c). 4.28 mg
,• (d). 10 hours
