Answer:
9 days before only 8 mg is still emitting radiation.
Step-by-step explanation:
The exponential model is:
[tex]y(t) = a(0.5)^{\frac{t}{28}}[/tex]
In which a is y(0), that is, the initial quantity.
10 mg is ingested by a patient
This means that [tex]a = 10[/tex], and thus:
[tex]y(t) = 10(0.5)^{\frac{t}{28}}[/tex]
How many days before only 8 mg is still emitting radiation?
This is t for which y(t) = 8. So
[tex]y(t) = 10(0.5)^{\frac{t}{28}}[/tex]
[tex]8 = 10(0.5)^{\frac{t}{28}}[/tex]
[tex](0.5)^{\frac{t}{28}} = \frac{8}{10}[/tex]
[tex](0.5)^{\frac{t}{28}} = 0.8[/tex]
[tex]\log{(0.5)^{\frac{t}{28}}} = \log{0.8}[/tex]
[tex](\frac{t}{28})\log{0.5} = \log{0.8}[/tex]
[tex]\frac{t}{28} = \frac{\log{0.8}}{\log{0.5}}[/tex]
[tex]t = 28\frac{\log{0.8}}{\log{0.5}}[/tex]
[tex]t = 9[/tex]
9 days before only 8 mg is still emitting radiation.
