Asked by yaz
A veterinarian has instructed Harrison to give his 75lb dog one 325 mg aspirin tablet for arthritis. the amount of aspirin A remaining in the dog's body after t minutes can be expressed by A=325(1/2)^(t/15). write and solve a logarithmic inequality to find the time it takes for the amount of aspirin to drop below 50 mg.
Answers
Answered by
Damon
50 = 325 * (.5)^(t/15)
0.1538 = .5^(t/15)
log (0.1538) = (t/15) log (.5)
-.8129= -.301 (t/15)
t = 15 ( 2.7)
t = 40.5
NOW, there is much more to this problem than meets the eye.
since it is (1/2)^t/something
when t is that something, it will be 1/2
in other words that something is the "half life" of the stuff (radioactive stuff or whatever)
That means every 15 minutes, the stuff is half gone, so in 30 minutes 1/4 is left and in 45 minutes 1/8 is left etc.
so as a quick check, after 45 minutes the should be 325/8 or about 41 left. That makes our 50 left after 40.5 minutes pretty reasonable.
They have sneaked a lot of physics into this little question :)
0.1538 = .5^(t/15)
log (0.1538) = (t/15) log (.5)
-.8129= -.301 (t/15)
t = 15 ( 2.7)
t = 40.5
NOW, there is much more to this problem than meets the eye.
since it is (1/2)^t/something
when t is that something, it will be 1/2
in other words that something is the "half life" of the stuff (radioactive stuff or whatever)
That means every 15 minutes, the stuff is half gone, so in 30 minutes 1/4 is left and in 45 minutes 1/8 is left etc.
so as a quick check, after 45 minutes the should be 325/8 or about 41 left. That makes our 50 left after 40.5 minutes pretty reasonable.
They have sneaked a lot of physics into this little question :)
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