anonymous
  • anonymous
75,000 live bacteria are present in a culture in a flask. When an antibiotic is added to the culture, the number of live bacteria is reduced by half every four hours. Approximately how many hours have passed when there are 2000 bacteria left alive? A. 16.2 hours B. 16.4 hours C. 16.0 hours D. 16.6 hours
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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campbell_st
  • campbell_st
This can be solved by thinking about the problem as a geometric sequence 75000, 37500, 18750, ... so you are looking the the value of n that makes \[2000 = 75000 \times (\frac{1}{2})^{n - 1}\] then \[\frac{2000}{75000} = (\frac{1}{2})^{n - 1}\] take the log of both sides and apply log laws \[\log(\frac{2000}{75000}) = (n - 1) \log(\frac{1}{2})\] now you are in a position to solve for n
anonymous
  • anonymous
you wanna just tell me the answer 😭
campbell_st
  • campbell_st
well Its interesting that the answer's don't match the question....

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campbell_st
  • campbell_st
here is why... |dw:1437005483474:dw|
campbell_st
  • campbell_st
if you solve for n, multiply the answer by 4, as each term in n represents 4 hours
mathstudent55
  • mathstudent55
@campbell_st Why do you have n - 1 as the exponent? I thought it should be n. At time 0, n = 0, \(75000(0.5)^n = 75000(0.5)^0 = 75000\)
anonymous
  • anonymous
@mathstudent55 how would I do this
campbell_st
  • campbell_st
@mathstudent55 I solved it but find the term in a geometric sequence.... \[a^n = a_{1} \times r^{n - 1}\] solve for n then multiply the answer by 4 as each n represents 4 hours
campbell_st
  • campbell_st
so it looks like it takes sometime between 20 and 24 hours for the population to reach approx 2000,
mathstudent55
  • mathstudent55
Since it takes 4 hours to decay to half, let's use the variable n to mean a period of 4 hours. At time = n = 0, there are 75,000 bacteria After 1 period of 4 hours, the bacteria becomes half, so at n = 1, there are 75,000(0.5) After 2 periods of 4 hours, at n = 2, there are 75,000(0.5 * 0.5 = 75,000 * 0.5^2 After 3 periods of 4 hours, at n = 3, there are 75,000 * 0.5^3 This sequence of calculations suggests that at the end of n periods of 4 hours, the bacteria population will be \(\large P = (75,000)0.5^n\) Since we are looking for when the population will be 2000, we substitute 2000 for P, and solve for n. Since our unknown, n, is an exponent, you need to use logarithms. \(2000 = (75,000)0.5^n\) Switch sides: \((75,000)0.5^n = 2000\) Divide both sides by 75,000 \(0.5^n = \dfrac{2}{75} \) Take the log of both sides: \(\log 0.5^n = \log \dfrac{2}{75} \) \(n \log 0.5 = \log 2 - \log 75\) \(n = \dfrac{\log 2 - \log 75}{\log 1 - \log 2} \) \(n = \dfrac{\log 2 - \log 75}{- \log 2} \) \(n = \dfrac{\log 75 - \log 2}{\log 2} \) \(n = 5.2288...\) Since n represents a period of 4 hours, multiply the value of n by 4 to obtain your answer.
campbell_st
  • campbell_st
so 4 x 5.22 isn't an answer choice...

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