If 900 grams of a radioactive substance are present initially and three years later only 300 grams remain, how much of the substance will be present after six years?

Answer 1

use y=ab^x

300=900b^3 (900 initially and after 3 years 300 remain)

divide both sides by 900

(1/3)=b^3

take the cube root of both sides

(1/3)^(1/3)=b

b=0.6933612744

after 6 years we have y=900(0.6933612744)^6

y=900*0.1111111111...

y=100 grams

note that 0.1111111111111... is equal to 1/9 and 900 * 1/9=100

Answer 2

Use the equation A(t) = A₀e^kt. Put in 900 for A₀ (beginning amount), 300 for A(t) (ending amount), 3 for t, and solve for k: 300 = 900e^3k; 1/3 = e^3k; ln(1/3) = 3k; k = -.366204, Now put k back in, and use 6 for t to solve back for A(t): A(t) = 900e^-.366204*6 =100.