Exponentialand Logarithmic Models

Question

You go to the doctor and he gives you 15 milligrams of radioactive dye. After 16 minutes, 6.75 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.

Answer 1

Every 16 minutes, the radioactivity decays by 6.75/15 = 0.45 of the original amount.

You need to wait for the amount remaining to be 2 mg.

So,

Final amount = initial amount * (0.45)^[(time in minutes)/(16 minutes)]

2 = 15 * (0.45)^(t/16)

Divide both sides of the above equation by 15:

2/15 = (0.45)^(t/16)

Take the log base 0.45 of both sides:

log _0.45 (2/15) = t/16

Multiply both sides by 16, and then use the change of base formula to complete the solution process:

t = 16 * log _0.45 (2/15) = 16 * [ ln (2/15) / ln (0.45) ]

t ≈ 40.37337208 minutes,

or t ≈ 41 minutes to make sure you are under 2 mg radioactivity.

Note: t = 40 minutes would be the answer rounded to the nearest minute, even though the radiation detectors would probably go off, since the radiation would be slightly over 2 mg.

Answer 2

A=15e^rt

6.75=15e^16r

divide by 15

6.75/15= e^16r

take natural logs of both sides

ln(6.75/15)= 16r

divide by 16

r= [ln(6.75)/15)]/16= about -.05= 5% per minute decay rate

to reach 2 mg

2= 15e^(ln(6.75/15)/16)t

solve for t

2/15= e^-.05t

ln(2/15)=- .05t

t= ln(2/15)/-.05= about 40.3 minutes

= 40 minutes, rounded to nearest minute

Answer 3

6.75 = 15e^16k

ln 6.75 = 16k ln 15, solve for k

2 = 15e^kt, substitute value of k from above and solve for t.