Growth of a Culture of Bacteria Day Number of bacteria per milliliter at end of day 1 2 3 A

Question

Growth of a Culture of Bacteria
Day	Number of bacteria per milliliter at end of day
1	$2 point 5 times 10 to the power 5$
2	$5 point 0 times 10 to the power 5$
3	$1 point 0 times 10 to the power 6$

Accepted Answer

Choice D is correct. The number of bacteria per milliliter is doubling each day. For example, from day 1 to day 2, the number of bacteria increased from 2.5 &times; 10 5 to 5.0 &times; 10 5 . At the end of day 3 there are 10 6 bacteria per milliliter. At the end of day 4, there will be 10 6 &times; 2 bacteria per milliliter. At the end of day 5, there will be (10 6 &times; 2) &times; 2, or 10 6 &times; (2 2 ) bacteria per milliliter, and so on. At the end of day d , the number of bacteria will be 10 6 &times; (2 d - 3 ). If the number of bacteria per milliliter will reach 5.12 &times; 10 8 at the end of day d , then the equation must hold. Since 5.12 &times; 10 8 can be rewritten as 512 &times; 10 6 , the equation is equivalent to . Rewriting 512 as 2 9 gives d - 3 = 9, so d = 12. The number of bacteria per milliliter would reach 5.12 &times; 10 8 at the end of day 12. Choices A, B, and C are incorrect. Given the growth rate of the bacteria, the number of bacteria will not reach 5.12 &times; 10 8 per milliliter by the end of any of these days.

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