If 6.7% is gone, then 93.3% remains.
You want a formula like A = (1/2)^(t/k)
(1/2)^(8.6/k) = 0.933
8.6/k log(1/2) = log0.933
8.6/k = log.933/log.5 = 0.1
k = 86
So the half-life is 86 hours.
Assume a quantity decreases by 6.7% in 8.6 hours.
What is the half-life of the substance? (in hours)
(Your answer should be accurate to two decimal places.)
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