hello, I found this intresting problem on a website. I did the problem but i'm pretty sure this is a tricky 1

Question

Find the half- life of a radio-active substance that is reduced by 20% in 30 hours. (Give your answer correct to three decimal places.)

Answer 1

In 30 hours, you have 4/5 remaining
in 60 hours, you have (4/5)^2 remaining (16/25
in t hours, you have (4/5)^t

(4/5)^t =1/2
Now solve for t
Take the Log or natural log(ln) of both sides
Log( (4/5)^t) = Log(1/2)

t Log(4/5) = Log(1/2)

t = Log(1/2) /Log(4/5) = -0.301029996 / -0.096910013 = 3.10628372
t =3.106 hours



EDIT----
gianlino(below) is correct. The answer I gave 3.106 hours should be multiplied by 30. I set the problem up as t = the number of hours, but calculated t as how many times 30 hours had passed.
So the time should have been 3.10628372*30 = 93.1885116
93.189 hours.

Answer 2

Yassine,
Atomic decay is an exponential process, whose rate of decrease is proportional to the current amount of the substance. See the Wikipedia source for a good description.

Answer 3

As you can see... It's way too "interesting" for me to even grasp at the concept. My guess is .007 / 80+30/24= ANSWER

Answer 4

Just multiply the prevous answer by 30...