A radioactive substance decays exponentially. A scientist begins with 100 milligrams of a radioactive substance. After 19 hours, 50 mg of the substance remains. mg

Answer 1

If the decay is exponential (as radioactive things do when they decay; we say this in terms of half-lives) and we let A(t) be the amount of the substance left t hours after first observed (that is, we're assuming that A(0)=100, and we're also given that A(19)=50), then A(t) should take the form

A(t)=[A(0)]*(1/2)^{kt}, where k is a constant we need to figure out (we're using 1/2 as the base instead of e; we could use any other base we want, but you're just going to have a much easier time using 1/2 since the two amounts you were given were 50 and 100). Let's figure k out: 50=A(19)=100*(1/2)^{19k}, i.e. (1/2)^{19k}=1/2, meaning that 19k=1, so k=1/19. Our formula for A(t) is then A(t)=100*(1/2)^{t/19}. If you want to find out how much is left at a given time, plug that time t into this formula. If you want to know at what time t A(t) hits a particular value B, plug that value B in as A(t) into our formula and solve for t by noting B/100=(1/2)^{t/19}, so then t/19=log_{1/2}(B/100)=log(B/100)/log(1/2) (if I write log without a base, it means log base 10; I can also use the natural log ln just as easily).