Question

homework assignment 6.7 exponential and logarithmic models
due friday by 11:59pm points 10 submitting an external tool
homework assignment 6.7 exponential and logarithmic models
score: 0/10 answered: 0/10
question 1
you go to the doctor and he gives you 13 milligrams of radioactive dye. after 16 minutes, 7.75 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.
question help: video read written example

Explanation:

Step1: Find the decay model

The general form of an exponential decay model is \(y = a\cdot e^{kt}\), where \(a\) is the initial amount, \(y\) is the amount at time \(t\), and \(k\) is the decay constant.
We know that \(a = 13\) (initial amount of dye), \(t = 16\) minutes, and \(y=7.75\).
Substitute these values into the equation: \(7.75=13\cdot e^{16k}\).
Divide both sides by \(13\): \(\frac{7.75}{13}=e^{16k}\), so \(0.59615 = e^{16k}\).
Take the natural logarithm of both sides: \(\ln(0.59615)=16k\).
Calculate \(\ln(0.59615)\approx - 0.517\), then \(k=\frac{-0.517}{16}\approx - 0.0323\).
So the model is \(y = 13\cdot e^{- 0.0323t}\).

Step2: Solve for \(t\) when \(y = 2\)

Set \(y = 2\) in the model: \(2=13\cdot e^{-0.0323t}\).
Divide both sides by \(13\): \(\frac{2}{13}=e^{-0.0323t}\), so \(0.15385 = e^{-0.0323t}\).
Take the natural logarithm of both sides: \(\ln(0.15385)=-0.0323t\).
Calculate \(\ln(0.15385)\approx - 1.879\).
Then \(t=\frac{-1.879}{-0.0323}\approx58.2\).

Answer:

\(58\)