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Applications of Logarithms & Exponentials

Applications of Logarithms & Exponentials. We have now reached a stage where trial and error is no longer required!. Ex1. Ex2. Solve lnx = 3.5 to 3 dec places. Solve e x = 14 to 2 dec places. *********. **********. e x = 14 . lnx = 3.5. lne x = ln14 . e lnx = e 3.5. x = e 3.5.

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Applications of Logarithms & Exponentials

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  1. Applications of Logarithms & Exponentials We have now reached a stage where trial and error is no longer required! Ex1 Ex2 Solve lnx = 3.5 to 3 dec places. Solve ex = 14 to 2 dec places. ********* ********** ex = 14 lnx = 3.5 lnex = ln14 elnx = e3.5 x = e3.5 x = ln14 x = 33.115 x = 2.64 Check ln33.115 = 3.4999…. Check e2.64 = 14.013

  2. Ex3 Solve 3x = 52 giving answer to 5 dec places. ********* 3x = 52 ln3x = ln52 xln3 = ln52 (law3)‏ x = ln52  ln3 x = 3.59658 Check: 33.59658 =52.0001….

  3. Ex4 A radioactive substance decays according to the formula At = A0e–0.55t where A0 is the initial quantity of the substance and At is the quantity remaining after t hours. Find the half-life of the substance to the nearest second. ****** For half-life we need At = 1/2A0 A0e–0.55t = 1/2A0 e–0.55t = 1/2 lne–0.55t = ln 1/2 -0.55t = ln1/2 t = ln1/2  (-0.55)‏ t = 1.260…..hrs ( 0.260… X 60 = 15.616..)‏

  4. t = 1hr 15.616..mins ( 0.616.. X 60 = 36.96..)‏ t = 1hr 15mins 37secs Ex5 The number of bacteria on an Agar plate grows according to the formula An = A0e0.75n where A0 is the initial number of bacteria and An is the number present after n hours. How long does it take for the number to increase “tenfold” ? *********** Tenfold is ten times as much !!

  5. For tenfold we need An = 10A0 A0e0.75n = 10A0 e0.75n = 10 lne0.75n = ln10 0.75n = ln10 n = ln10  0.75 n = 3.07..hrs ( 0.07.. X 60 = 4.206…) n = 3hrs 4.206..mins (0.206.. X 60 = 12.4..) n = 3hrs 4mins 12secs Check: If A0 = 5 then A3.07 = 5 X e (0.75 x 3.07) = 5 X e 2.3025 = 49.9957….  50

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