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homework assignment 6.7 exponential and logarithmic models
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homework assignment 6.7 exponential and logarithmic models
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you go to the doctor and he gives you 13 milligrams of radioactive dye. after 20 minutes, 5.25 milligrams of
dye remain in your system. to leave the doctors office, you must pass through a radiation detector without
sounding the alarm.
if the detector will sound the alarm if more than 2 milligrams of the dye are in your system, how long will
your visit to the doctor take, assuming you were given the dye as soon as you arrived?
give your answer to the nearest minute.
you will spend minutes at the doctors office.
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Explanation:

Step1: Find the decay constant \(k\)

The general formula for exponential decay is \(A(t)=A_0e^{-kt}\), where \(A_0 = 13\) (initial amount), \(A(20)=5.25\) (amount after \(t = 20\) minutes).
Substitute into the formula: \(5.25=13e^{-20k}\).
Divide both sides by \(13\): \(\frac{5.25}{13}=e^{-20k}\).
Take the natural logarithm of both sides: \(\ln(\frac{5.25}{13})=-20k\).
Solve for \(k\): \(k =-\frac{\ln(\frac{5.25}{13})}{20}\approx-\frac{\ln(0.4038)}{20}\approx\frac{-(-0.907)}{20}= 0.04535\).

Step2: Find the time \(t\) when \(A(t) = 2\)

Use the formula \(A(t)=A_0e^{-kt}\) with \(A_0 = 13\), \(k = 0.04535\), and \(A(t)=2\).
So, \(2 = 13e^{-0.04535t}\).
Divide both sides by \(13\): \(\frac{2}{13}=e^{-0.04535t}\).
Take the natural logarithm of both sides: \(\ln(\frac{2}{13})=-0.04535t\).
Solve for \(t\): \(t=\frac{\ln(\frac{2}{13})}{- 0.04535}\approx\frac{-1.771}{-0.04535}\approx39\).

Answer:

\(39\)