Thread: Applications of Exponential and Logarithmic Functions

Hey guys, this is my first post and I would like to be active on these forums and help others where I can.

I ran into some problems that I couldn't find a solution to. There are three questions I'm struggling with and there all somewhat related in a sense.

1.The amount of particulate matter left in solution during a filtering process is given by the equation where n is the number of filtering steps. Find the amounts left for and . (Round to the nearest whole number.)

-This one makes no sense to me and I tried plugging in the n values. maybe I am suppose to be taking the log of the equation??

2. How long will it take a sample of radioactive substance to decay to half of its original amount, if it decays according to the function A(t) = 200e-0.131t, where t is the time in years? Round to the nearest hundredth year.

-This question I feel like I don't know what to plug in or how to go through finding the correct solution.


3. Use the formula N = Iekt, where N is the number of items in terms of the initial population I, at time t, and k is the growth constant equal to the percent of growth per unit of time. A certain radioactive isotope has a half-life of approximately 1750 years. How many years would be required for a given amount of this isotope to decay to 20% of that amount?

- I tried setting this up in this way. ln(.5) = 1750 this came out to .000120547(can I stop after the 0 and leave it as .00012? Not sure If I'm doing this one right either and these three questions didn't seem like rest of them and I can't figure out!

You haven't included an equation for Question 1, so we can't do anything with it...

For Question 2, do you know how much substance there originally was before decay started? If not, how are we supposed to know what half of its amount is?

Your exactly right. That is why I am having problems with these questions. That is all of the information I am given. I have listed exactly what the problem states. There is no other information I have.