1. Finding the equation of an exponential curve using ln Plot the points and verify that the curve is exponential xy tCaffeine 0126 1111 298 387 477 569 661 754 847 942 1038 1134 1230 1326 1423 1521 1618 1716 1814 1913 2011 2110 229 238 247 The table shows the amount of caffeine in the bloodstream after a time t mins

2. The equation is N = N 0 e -bt using the exponential function e N = Number present after a time period t N 0 = Number present initially (time t = 0) t = units of time b = constant

3. A B Taking lns of both sides ln N = ln(N 0 e bt ) ln N = ln N 0 + lne bt Using the addition rule ln(AB) = lnA + lnB ln N = ln N 0 + lne Using the drop down infront rule. ln N = ln N 0 + bt lne = 1 ln N = bt + lnN 0 Rearranging to match with y=mx + c The equation is N = N 0 e bt bt

4. So make a new table of values x = t y = lnN ln N = b t + lnN 0 Matching up with y=mx + c : y = ln N gradient = b x = t C = ln N 0

5. Plot t values on the x axis and lnN values on the y axis tCaffeine Nln N 0126 4.83628 1111 4.70953 298 4.58497 387 4.46591 477 4.34381 569 4.23411 661 4.11087 754 3.98898 847 3.85015 942 3.73767 1038 3.63759 1134 3.52636 1230 3.4012 1326 3.2581 1423 3.13549 1521 3.04452 1618 2.89037 1716 2.77259 1814 2.63906 1913 2.56495 2011 2.3979 2110 2.30259 229 2.19722 238 2.07944 247 1.94591

6. y = -0.1202x + 4.8292 The equation of the line is ln N = b t + lnN 0 gradient = b = -0.1202 so b = -0.1202 Matching up : C = ln N 0 = 4.8292 To find N 0 use forwards and back N 0 = e 4.8282 =125.11 N 0  ln it = 4.8292 4.8292  e it = N 0

7. The exponential equation is N = N 0 e bt N = 125.11×e -0.1202t We do not need to find out how long for the caffeine to ½.

8. Using the Equation The exponential curve for the data earlier was given by N = 125.11e -0.1202t. So if the time t is given it is easy to work out the amount of caffeine. If t = 12mins then N = 125.11e -0.1202  12 = 29.6mg Replace t by the time required

9. But if the caffeine N is given then we have to use lns to solve for t. Find the time to reach 50mg N = 125.11e -0.1202t t  ×–0.1202  e it  ×125.11 = 50 50  ÷125.11  ln it  ÷–0.1202 50 = 125.11e -0.1202  t Using lns Forwards Backwards