Asked by Zyere
The half-life of a specific radioactive compound is 20 years. The equation used to model this half-life is y = 5 1/2(t/20) where t represents time. Currently, there are 5 pounds of this compound 15 feet below the earth's surface. When will there only be 4 pounds of this compound left? Round your answer to the nearest tenth of a year.
Answers
Answered by
Reiny
You typed the equation incorrectly, it should say
y = 5 (1/2)^(t/20)
so ...
4 = 5(1/2)^(t/20)
.8 = (.5)^(t/20)
take log of both sides
log.8 = log(.5^(t/20))
log.8 = (t/20)log.5
log.8/log.5 = t/20
t = 20(log.8)/log.5)
I leave the button-pushing up to you
y = 5 (1/2)^(t/20)
so ...
4 = 5(1/2)^(t/20)
.8 = (.5)^(t/20)
take log of both sides
log.8 = log(.5^(t/20))
log.8 = (t/20)log.5
log.8/log.5 = t/20
t = 20(log.8)/log.5)
I leave the button-pushing up to you
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