Plugging in 10000 for t and solving for A we get: : A certain radioactive isotope element decays exponentially according to the model , where A is the number of grams of the isotope at the end of t days and Ao is the number of grams present initially. If we are looking for the half-life of this isotope, what variable are we seeking? It looks like we don’t have any values to plug into A or Ao.

However, the **problem** did say that we were interested in the HALF-life, which would mean ½ of the initial amount (Ao) would be present at the end (A) of that time. Replacing A with .5 Ao and solving for t we get: : Prehistoric cave paintings were discovered in a cave in Egypt. Using the exponential decay model for **carbon**-14, , estimate the age of the paintings.

The decay model describes the amount of *carbon*-14 present after t years.

**How** many grams of **carbon**-14 will be present in this artifact after 10,000 years?

The diagram below shows exponential growth:: The exponential growth model describes the population of a city in the United States, in thousands, t years after 1994.

The half-life of a given substance is the time required for half of that substance to decay or disintegrate.

The diagram below shows exponential decay:: An artifact originally had 12 grams of **carbon**-14 present.

The reason I showed you using the formula was to get you use to it.

Just note that when it is the initial year, t is 0, so you will have e raised to the 0 power which means it will simplify to be 1 and you are left with whatever Ao is. Well, k = .0198026, so converting that to percent we get 1.98026% for our answer.

The diagram below shows exponential growth:: The exponential growth model describes the population of a city in the United States, in thousands, t years after 1994.The half-life of a given substance is the time required for half of that substance to decay or disintegrate.The diagram below shows exponential decay:: An artifact originally had 12 grams of **carbon**-14 present.The reason I showed you using the formula was to get you use to it.Just note that when it is the initial year, t is 0, so you will have e raised to the 0 power which means it will simplify to be 1 and you are left with whatever Ao is. Well, k = .0198026, so converting that to percent we get 1.98026% for our answer.If we are looking for the number of grams of *carbon*-14 present, what variable do we need to find? What are we going to plug in for t in this *problem*?