Question

A patient takes a medication with a half life. initially, there are 11 milligrams of the medication in the patient's system. after 70 minutes there are 7 milligrams. after how many minutes will there be only 3 milligrams remaining in the patient's system? round your answer to the nearest whole number, and do not include units.

Answer 1

Using an exponential function, it is found that there will be only 3 milligrams remaining in the patient's system after 201 minutes.

What is an exponential function?

A decaying exponential function is modeled by:

$A(t) = A(0)(1 - r)^t$

In which:

• A(0) is the initial value.

• r is the decay rate, as a decimal.

In this problem, initially, there are 11 milligrams on the patient's system, hence A(0) = 11. After 70 minutes there are 7 milligrams, hence A(70) = 7, and this is used to find r.

$A(t) = A(0)(1 - r)^t$

$7 = 11(1 - r)^{70}$

$(1 - r)^{70} = \frac{7}{11}$

$\sqrt[70]{(1 - r)^{70}} = \sqrt[70]{\frac{7}{11}}$

$1 - r = \left(\frac{7}{11}\right)^\frac{1}{70}$

1 - r = 0.99356387084

r = 1 - 0.99356387084

r = 0.00643612916

Hence the equation for the amount after t minutes is:

$A(t) = 11(0.99356387084)^t$

In will be of 3 mg when A(t) = 3, hence:

$A(t) = 11(0.99356387084)^t$

$3 = 11(0.99356387084)^t$

$(0.99356387084)^t = \frac{3}{11}$

$\log{(0.99356387084)^t} = \log{\left(\frac{3}{11}\right)}$

$t\log{(0.99356387084)} = \log{\left(\frac{3}{11}\right)}$

$t = \frac{\log{\left(\frac{3}{11}\right)}}{\log{(0.99356387084)}}$

t = 201

There will be only 3 milligrams remaining in the patient's system after 201 minutes.

More can be learned about exponential functions at brainly.com/question/25537936