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 × 10⁵ to 5.0 × 10⁵. At the end of day 3 there are 10⁶ bacteria per milliliter. At the end of day 4, there will be 10⁶ × 2 bacteria per milliliter. At the end of day 5, there will be (10⁶× 2) × 2, or 10⁶ × (2²) bacteria per milliliter, and so on. At the end of day d, the number of bacteria will be 10⁶× (2^{d – 3}). If the number of bacteria per milliliter will reach 5.12 × 10⁸ at the end of day d, then the equation $10 to the power 6, end power, times 2 to the power d minus 3, end power, equals 5 point 1 2, times 10 to the power 8$ must hold. Since 5.12 × 10⁸ can be rewritten as 512 × 10⁶, the equation is equivalent to $2 to the power d minus 3, end power, equals 512$ . Rewriting 512 as 2⁹ gives d – 3 = 9, so d = 12. The number of bacteria per milliliter would reach 5.12 × 10⁸ 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 × 10⁸ per milliliter by the end of any of these days.

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Growth of a Culture of Bacteria Day Number of bacteria per milliliter at end of day

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