A sample originally contains 24 grams of a radioactive isotope. After 18 days, 18.8 grams of the isotope remains in the sample.
What is the half-life of the isotope?
+128460 We have that
18.8 = 24 b^18 divide both sides by 24
18.8 / 24 = b^18 take the 18th root of both sides
(18.8/24)^(1/18) = b
So....we want to find this [the half-life is the tme when 12 g remain ]
12 = 24 [ (18.8.24)^(1/18)]^(t) where t is the number of days we seek
12 = 24 [ 18/8/24]^(t/18)
Divide both sides by 24
(1/2) = (18.8/24)^(t/18) take the log of both sides
log (1/2) = log (18.8/24)^(t/18) and we can write
log (1/2) = (t/18) log (18.8/24) isolate t
18 log (1/2) / log (18.8 / 24) = t ≈ 51.1 days
