Science:Math Exam Resources/Courses/MATH100/December 2015/Question 06 (a)
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Question 06 (a) 

A colony of bacteria doubles every 4 hours. If the colony has 2000 cells after 6 hours, how many cells were present initially? Simplify your answer. 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! 
Hint 

The number of the cells in hours has the form of for some constant and . 
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Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. The number of cells in hours obeys the rule: for some and , where is the initial number of cells at , so the question asks for . Since a colony of bacteria doubles every 4 hours, the number of cells in 4 hours doubles the initial number of the cells:
On the other hand, the number of cells in 6 hours is , so that we have . Therefore, . 