Doubling Time - Bacteria Growth

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khan

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A biology laboratory starts a Petri dish with 1000 bacteria. After 30.0 hours, the population is 7250.
Determine the doubling time of the bacteria to the nearest tenth of an hour. The doubling-time equation
is
A=Ai (2)^t/d

where A represents the population after a period of time, Ai represents the initial population, t
represents the time, and d represents the time it takes for the population to double.

Could anyone please help me solve this problem. Thank in advance.

the equation should be: 7250=1000(2)^30/d
dividing both sides by 1000 it would be 72.5=2^30/d

I couldn't solve it for d.

Any help would be much appreciated.
 
You should use parentheses: 72.5=2^(30/d)

I would now take the log of both sides (using whatever base you prefer).
 
I tried to do it like this:
Log 72.5 = Log 2. 30/d

But I don't know how to solve it to get the value of d.
 
So you have not learned the "properties of logarithms" (which are the whole point of using them!)? The one you need here is that \(\displaystyle log(a^b)= b log(a)\) for any numbers a and b. So \(\displaystyle log(72.5)= log(2^{30/d)= (30/d)log(2)\).
 
Hi, thanks so much everyone for your help. Actually the laws of Logarithms are in the next chapter in my course, but after reviewing those laws, I was able to solve the question. Cheers!
 
You could solve the equation 72.5=2^(30/d) without using those laws, if you have learned to rewrite an exponential equation as a logarithmic equation, and if you have a way to find base 2 logs. It becomes log_2(72.5) = 30/d, and you could evaluate the left side and then solve for d. Otherwise, I am curious how they expected you to solve it; if I were working with you face to face, I would have asked to look through your book or notes to see what else you have learned.
 
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