Question

21 mark for review the amount of a certain medicine, in milligrams, in a patients body t hours after an initial dose can be modeled by the exponential decay function f given by f(t)=ab^t. an initial dose of the medicine is 10 milligrams, and after 2 hours the amount in the body is 5 milligrams. at what time t will the amount in the body be 0.01 milligram? (assume no additional doses of the medicine are given after the initial dose.) a 3.996 b 4.983 c 9.966 d 19.932

Explanation:

Step1: Find the value of \(a\)

When \(t = 0\), \(f(0)=ab^{0}=a\). Since the initial dose is 10 milligrams, \(a = 10\). So the function is \(f(t)=10b^{t}\).

Step2: Find the value of \(b\)

When \(t = 2\), \(f(2)=10b^{2}\). Given \(f(2)=5\), we have \(10b^{2}=5\). Then \(b^{2}=\frac{5}{10}=\frac{1}{2}\), so \(b=\frac{1}{\sqrt{2}}\) (because it's an exponential decay function, \(0 < b<1\)).

Step3: Solve for \(t\) when \(f(t)=0.01\)

Set up the equation \(10\times(\frac{1}{\sqrt{2}})^{t}=0.01\). First, rewrite it as \((\frac{1}{\sqrt{2}})^{t}=\frac{0.01}{10}=0.001\).
Take the natural logarithm of both sides: \(\ln((\frac{1}{\sqrt{2}})^{t})=\ln(0.001)\).
Using the property \(\ln(x^{y}) = y\ln(x)\), we get \(t\ln(\frac{1}{\sqrt{2}})=\ln(0.001)\).
Since \(\ln(\frac{1}{\sqrt{2}})=-\frac{1}{2}\ln(2)\approx - 0.3466\) and \(\ln(0.001)=-6.9078\).
Then \(t=\frac{\ln(0.001)}{\ln(\frac{1}{\sqrt{2}})}=\frac{- 6.9078}{-0.3466}\approx9.966\).

Answer:

C. 9.966