Finding doubling time with messy numbers
Recall that we started out with an equation:
N_{t} = N_{0} 2^{g}
Let's say I told you that N_{0} was 10 million and N_{t} was 80 million one hour later. You would probably think to yourself, "well, the population doubled three times, from 10 to 20 million, from 20 to 40 million, and from 40 to 80 million. Since three doublings took 1 hour, that means each doubling took 20 minutes."
Another somewhat fancier way of getting the same answer would be to substitute the information that you know (N_{t} and N_{0}) into the equation above, and solve for the information that you don't know (number of generations, g). Here's how that would look:
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What is the doubling time?
(10 million > 80 million in 1 hour)
Hint  Explanation  
sub in the known info  80 million = 10 million * 2^{g}  
divide out the initial population  8 = 2^{g}  
use common sense!  g must be 3  
3 doublings in 60 minutes means...  60min/3doublings = 20min/doubling, so the doubling time is 20 min 
We will need to do same basic procedure with messier numbers. Let's say the population increased from 10 to 90 million in one hour.
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What is the doubling time?
(10 million > 70 million in 1 hour)
Hint  Explanation  
sub in the known info  70 million = 10 million * 2^{g}  
divide out the initial population  7 = 2^{g}  
use common sense!  g must be ... hmmm...  
use exponent rules and a calculator  log(7) = g * log(2), so g = log(7)/log(2) = 2.81 

more calculator  60min/2.81doublings = 21.4min/doubling, so the doubling time is 21.4 min 
In a word: when you have a problem with exponents, try logs. By now you should have seen logs in several of these modules, but if you don't remember the basics, you can click here for a refresher. Remember, logs are a biologist's best friend.
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