Question

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 doctor’s 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 doctor’s office.
question help: video read written example

Explanation:

Step1: Find the decay formula

The general formula for exponential decay is \(y = a\cdot b^{t}\), where \(a\) is the initial amount, \(b\) is the base of the exponential function, and \(t\) is the time.
We know that \(a = 13\), when \(t = 20\), \(y=5.25\). So \(5.25=13\cdot b^{20}\). Then \(b^{20}=\frac{5.25}{13}=\frac{21}{52}\), and \(b = (\frac{21}{52})^{\frac{1}{20}}\).

Step2: Set up the equation for \(y = 2\)

We want to find \(t\) when \(y = 2\). Using the formula \(y = 13\cdot b^{t}\), substitute \(y = 2\) and \(b = (\frac{21}{52})^{\frac{1}{20}}\) into it. So \(2=13\cdot(\frac{21}{52})^{\frac{t}{20}}\).

First, rewrite the equation as \((\frac{21}{52})^{\frac{t}{20}}=\frac{2}{13}\).

Take the natural logarithm of both sides: \(\ln((\frac{21}{52})^{\frac{t}{20}})=\ln(\frac{2}{13})\).

Using the property of logarithms \(\ln(a^{m})=m\ln(a)\), we get \(\frac{t}{20}\ln(\frac{21}{52})=\ln(\frac{2}{13})\).

Then \(t=\frac{20\ln(\frac{2}{13})}{\ln(\frac{21}{52})}\).

Calculate \(\ln(\frac{2}{13})\approx\ln(2)-\ln(13)\approx0.693 - 2.565=- 1.872\), \(\ln(\frac{21}{52})\approx\ln(21)-\ln(52)\approx2.996 - 3.951=-0.955\).

So \(t=\frac{20\times(- 1.872)}{-0.955}=\frac{-37.44}{-0.955}\approx39.2\).

Answer:

\(39\)