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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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
you wanna just tell me the answer 😭
well Its interesting that the answer's don't match the question....

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

here is why... |dw:1437005483474:dw|
if you solve for n, multiply the answer by 4, as each term in n represents 4 hours
@campbell_st Why do you have n - 1 as the exponent? I thought it should be n. At time 0, n = 0, \(75000(0.5)^n = 75000(0.5)^0 = 75000\)
@mathstudent55 how would I do this
@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
so it looks like it takes sometime between 20 and 24 hours for the population to reach approx 2000,
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.
so 4 x 5.22 isn't an answer choice...

Not the answer you are looking for?

Search for more explanations.

Ask your own question