Calculus help! Two years ago there were8 grams of a radioactive substance. Now there are 6 grams. How much....

Two years ago there were 8 grams of a radioactive substance. Now there are 6 grams. How much will remain 3 years from now?

The equation for radioactive decay is

A(t) = A(0)e^(kt)

where A(t) is the amount after t years, A(0) is the initial amount, k is the rate constant, and t is the time in years. We have

6 = 8e^(2k)

0.75 = e^(2k)

ln(0.75) = 2k

k = ln(0.75)/2 = -0.287682072/2 = -0.143841036

Our equation is now

A(t) = A(0)e^(-0.143841036t)

A(3) = 6e^[(-0.143841036)(3)]

= 6e^(-0.431523109)

= 6(0.649519053)

= 3.897114317 grams

within 2 years 2 grams of radioactive substance has disappeared.

this means that every year there will disappear 2/2=1 gram of radioactive substance.

so in 3 year time there will disappear 3*1=3 grams of radioactive substance.

so in 3 years from now there will remain 6-3=3 grams of radioactive substance

6 = 8 * 1 / 2^h ... where h is the number of half lives

2^h = 4 / 3

ln(2) h = ln(4/3)

h = ln(4/3) / ln(2) ~ 0.415 <=== 2 years = 0.415 half lives

3 years = 3/2 * 0.415 = 0.6225 half lives

Amount = 6 * 1 / 2^(0.6225) = 3.897 grams

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6 = 8e^(2k)

e^(2k) = 3/4

2k = ln(3/4)

k = ln(3/4) / 2

Amount = 6 e^(3 * ln(3/4) / 2)

Amount = 9 sqrt(3) / 4 ~ 3.897 grams

If it is lineal (Time, substance ) = ( 0,8) and now (2,6)
S-8= (6-8) /(2-0) (T-0)

S(T) = -T+8

3 years from now is 5 years from T=0
S(5) = -5+8= 3 grams