Solving Exponential Growth Problems

We will still be using the same formula we did to answer the questions above, we will just be using it to find a different variable.

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As mentioned above, in the general growth formula, k is a constant that represents the growth rate. Since we are looking for the population, what variable are we finding? What are we going to plug in for t in this problem?

Our initial year is 1994, and since t represents years after 1994, we can get t from 2005 - 1994, which would be 11.

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.

For example, bacteria will continue to grow over a 24 hours period, producing new bacteria which will also grow.

The bacteria do not wait until the end of the 24 hours, and then all reproduce at once.

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.

Plugging in 11 for t and solving for A we get: Looks like we have a little twist here.

Now we are given the population and we need to first find t to find out how many years after 1994 we are talking about and then convert that knowledge into the actual year.

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