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GROWTH & DECAY NB Removing 15% leaves behind 85% or 0.85 which is called the DECAY factor. Adding on 21% gives us 121% or 1.21 and this is called the GROWTH factor. Growth and decay factors allow us a quick method of tackling repeated % changes.

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Ex1An agar plate contains 10000 bacteria which are being killed off at a rate of 17% per hour by a particular disinfectant. (a) How many bacteria are left after 3 hours? (b) How many full hours are needed so that there are fewer than 4000 bacteria? Suppose that u n represents the number of bacteria remaining after n hours. Removing 17% leaves behind 83% so the DECAY factor is 0.83 and u n+1 = 0.83 u n *******************************

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(a)u 0 = 10000 u 1 = 0.83u 0 = 0.83 X 10000 = 8300 u 2 = 0.83u 1 = 0.83 X 8300 = 6889 u 3 = 0.83u 2 = 0.83 X 6889 = 5718 So there are 5718 bacteria after 3 hours. (b) u 4 = 0.83u 3 = 0.83 X 5718 = 4746 u 5 = 0.83u 4 = 0.83 X 4746 = 3939 This is less than 4000 so it takes 5 full hours to fall below 4000.

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Ex2The population of a town is growing at a rate of 14% per annum. If P 0 is the initial population and P n is the population after n years ….. then find a formula for P n in terms of P 0. Find roughly how long it takes the population to treble. *************************** Adding on 14% gives us 114% so the GROWTH factor is 1.14 and P n+1 = 1.14 P n

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P 1 = 1.14 P 0 P 2 = 1.14 P 1 = 1.14 X 1.14 P 0 = (1.14) 2 P 0 P 3 = 1.14 P 2 = 1.14 X (1.14) 2 P 0 = (1.14) 3 P 0 etc So in general we have P n = (1.14) n P 0 If the population trebles then we need to have P n > 3 P 0 or (1.14) n P 0 > 3 P 0 Dividing by P 0 we get (1.14) n > 3

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We now use a bit of trial and error along with the ^ or x y buttons on the calculator. If n = 5 then (1.14) 5 = 1.92… too small If n = 9 then (1.14) 9 = 3.25… too big If n = 7 then (1.14) 7 = 2.50… too small If n = 8 then (1.14) 8 = 2.85… too small but closest to 3. From the above we can say it takes just over 8 years for the population to treble.

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