You can put this solution on YOUR website! Find the half life of a radioactive substance that decays by 20% in 10 years.
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A(t)= a*b^t
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Given:
0.20a = ab^10
b^10 = 0.20
b = 0.2^(1/10)
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Find half-life:
(1/2)a = a*b^t
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Cancel the "a's" and substitute for "b":
1/2 = 0.2^(t/10)
Solve for "t":
(t/10)log(0.2) = log(0.5)
t/10 = log(0.5)/log(0.2)
t/10 = 0.4307
t = 4.307 years (half-life of the substance)
Cheers,
Stan H.
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