SOLUTION: A radioactive isotope has a half-life of 3000 years. If a sample of this isotope origanally has a mass 30g, what equation would model the mass of this sample over time? What would

Let A(t) represent the amount of 
isotope at time t. The formula is:

A(t) = Pert

At the start, when time = t = 0, 
then A(0) = 30

So we substitute 0 for t:

A(0) = Per*0

and we substitute 30 for A(0)

  30 = Pe0
  30 = P(1)
  30 = P

So now the formula

A(t) = Pert

becomes

A(t) = 30ert

Nw we read >>...half-life of 3000 years...<<

which means that in 3000 years the isotope will
have reduced to half of its original mass of 
30 grams.  That means that A(3000) = 15 grams.

So we substitute 3000 for t

A(t) = 30ert
A(3000) = 30er(3000)

and then substitute 15 for A(3000)

A(3000) = 30er(3000)

15 = 30er(3000)

Divide both sides by 30

 = e3000r

0.5 = e3000r

Taking the natural log of both sides,
we have 

ln(0.5) = 3000r

Use a calculato to get the left side:

-.6931471806 = 3000r

Divide both sides by 3000

-.00023104906 = r

So the formula

A(t) = 30ert

becomes

A(t) = 30e-0.00023104906t
 
That is the equation that models the mass,
which was the first part of your problem. 

Now for the second part, we only need to plug 
in 5 hours.  But we must change that to
years.  So

Change 5 hours to days by dividing by 24, getting
5 hours = 0.2083333333 days,

Now divide that by 365.25 to change it to years:

5 hours = 0.0005703855807 years

So plug that into

A(t) = 30e-.00023104906t

A(0.0005703855807) = 30e-.00023104906(0.0005703855807)

A(0.0005703855807) = 29.99999605 grams.

As we might guess, if it takes 3000 years for
30 grams to decrease to 15 grams, we wouldn't
expect it to have noticeably decreased from the
original 30 grams after only 5 hours!

Edwin