Question 1130
<pre><font face = "courier new">If 23.00 grams of a 45.00 gram sample of unknown isotope remain after 
17 days 12 hours, what is the half-life of the unknown isotope to the 
nearest minute?
The formula is 
               A = Pe<sup>rt</sup>
When time t=0 days, the amount A = 45g.  Substitute these two values
              45 = P<sup>er(0)</sup>
              45 = Pe<sup>0</sup>
              45 = P(1)
               P = 45 grams
so substitute 45 for P in A = Pe<sup>rt</sup>
               A = 45ert
When time t=17.5 days, the amount A = 23g.  Substitute these two values:
              23 = 45e<sup>r(17.5)</sup>
              23 = 45e<sup>17.5r</sup>
Divide both sides by 45
           23/45 = e<sup>17.5r</sup>
     .5111111111 = e<sup>17.5r</sup>
Rewrite this equation using the fact that
" N = e<sup>M</sup> can be rewritten as M = ln(N) "
           17.5r = ln(.5111111111)
Use a calculator to find ln(.5111111111) = -.6711682738
           17.5r = -.6711682738
Solve for r
               r = -.6711682738/17.5
               r = -.0383524728
So substitute this value of r in A = 45e<sup>rt</sup>
               A = 45e<sup>-.0383524728t</sup>
We wish to know how many minutes is required for it to become half
its original amount, or 1/2 of 45g, or 22.5 g.  So substitute this 
for A in A = 45e<sup>-.0394804867t</sup>
            22.5 = 45e<sup>-.0383524728t</sup>
Divide both sides by 45
         22.5/45 = e<sup>-.0383524728t</sup>
              .5 = e<sup>-.0383524728t</sup>
Rewrite this as 
          ln(.5) = -.0383524728t
Solve for t
    -.6931471806 = -.0383524728t
   -.0383524728t = -.6931471806
               t = -.6931471806/(-.0383524728)
               t = 18.07307665 days
To find hours, multiply .07307665 by 24
       18 days 1.753839658 hours
To find minutes, multiply .753839658 by 60
       18 days 1 hour 45.23037947 minutes or to nearest minute:
       18 days 1 hour, 45 minutes.
Edwin