Text Book Maths Problem - 3

Question

A substance is decaying such that the amount present (S grams) at any time (t days) is given by . S=200e^kt

 

Find k if there is 100 g present after 10 days;
 

Calculate how much is present after 18 days.

Answer 1

Jarrod,

 

There are two parts to this question, first we need to find k so let's start there.

 

We are given the equation S=200e^kt and S is the amount left at the end of an amount of time and t is time.

 

The first part of the problem tells us that there are 100g left after 10 days, so S=100 and t=10. Putting these into the equation we have

 

                                                       S = 200 e^kt

 

                                                       100 = 200 e^{k*10                              divide by 200 on both sides}

 

                                                       1/2 = e^{k*10                                       use ln to undo e and isolate k}

 

                                                       ln(1/2) = ln(e^10k)

 

                                                       -0.693 = 10k                           ^{divide by 10 on both sides}

 

                                                       -0.0693 = k

 

So the new equation with k is S = 200e^-0.0693t . The second part of the equation asks for us to find S when t = 18, we will use the same value for k (in most math and science problems, k is a consent and will remain the same for the problem unless you are told otherwise.) Lets put T into the equation and solve this part.

 

S = 200e^-0.0693t

 

S = 200e^{-0.0693*18                  Solve the exponent first}

S = 200e^-1.2474                        

S = 200*0.2873

S = 57.45

 

So there are 57.45g remaining after 18 days (remember that  S is in grams, so you should include the units in your answer.)

 

Let me know if you need any more help with this.