Since the time of the half-life is 3 hours, then we have to divide the number of hours given by 3
The form of the exponential function is
[tex]D(t)=a(b)^{\frac{t}{n}}[/tex]a is the initial amount
b is the factor of increasing or decreasing
In our situation:
Half-life means b = 1/2
Since the initial amount given is 20 mg, then
a = 20
Since the time of half-life is 3 hours, then
t must be the time divided by 3
The function which represents the situation is
[tex]D(t)=20(\frac{1}{2})^{\frac{t}{3}}[/tex]Since t = 3, then
[tex]\begin{gathered} f(3)=20(\frac{1}{2})^{\frac{3}{3}} \\ f(3)=20(\frac{1}{2})^1 \\ f(3)=10\text{ mg} \end{gathered}[/tex]a.
At t = 6
[tex]\begin{gathered} f(\frac{6}{3})=20(\frac{1}{2})^{\frac{6}{3}} \\ f(2)=20(\frac{1}{2})^2 \\ f(2)=20(\frac{1}{4}) \\ f(2)=5\text{ mg} \end{gathered}[/tex]At t = 9
[tex]\begin{gathered} f(9)=20(\frac{1}{2})^{\frac{9}{3}} \\ f(9)=20(\frac{1}{2})^3 \\ f(9)=20(\frac{1}{8}) \\ f(9)=2.5\text{ mg} \end{gathered}[/tex]b.
To find the amount at t = 10 hours, substitute t by 10
[tex]\begin{gathered} f(10)=20(\frac{1}{2})^{\frac{10}{3}} \\ f(10)=1.98425\text{ mg} \end{gathered}[/tex]We know that by using the exponential function above
2.
The function form is
[tex]D(t)=A(\frac{1}{2})^{\frac{t}{n}}[/tex]Where A = 20 -------- initial amount
n = 3 ------- the period of half-life
The formula is
[tex]D(t)=20(\frac{1}{2})^{\frac{t}{3}}[/tex]3.
The drug remains after 1 hour means substitute t by 1 first, then divide the answer by the initial amount, and change it to percent
[tex]\begin{gathered} D(1)=20(\frac{1}{2})^{\frac{1}{3}} \\ D(1)=15.87401052 \end{gathered}[/tex]We will find the percent
[tex]\begin{gathered} \text{ \%D=}\frac{15.87401052}{20}\times100\text{ \%} \\ \text{ \%D=79.37\%} \end{gathered}[/tex]To find the percent of the amount eliminated subtract 79.37% from 100%
[tex]\text{ \%E=100\%-79.37=20.63\%}[/tex]4.
The direction for adults is
Do not exceed 4 doses per 24 hours
5.
Since the table has a period of 2 hours, then we will use t = 2, 4, 6, 8, 10, 12 in the formula above to find the amount of Dex.
[tex]\begin{gathered} D(2)=20(\frac{1}{2})^{\frac{2}{3}}=12.599\text{ mg} \\ D(4)=20(\frac{1}{2})^{\frac{4}{3}}=7.937\text{ mg} \end{gathered}[/tex][tex]\begin{gathered} D(6)=20(\frac{1}{2})^{\frac{6}{3}}=5\text{ mg} \\ D(8)=20(\frac{1}{2})^{\frac{8}{3}}=3.150\text{ mg} \end{gathered}[/tex][tex]\begin{gathered} D(10)=20(\frac{1}{2})^{\frac{10}{3}}=1.984\text{ mg} \\ D(12)=20(\frac{1}{2})^{\frac{12}{3}}=1.25\text{ mg} \end{gathered}[/tex]