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Given that the initial population, $N_0$ of bacteria is $12$ and the population doubles every $20$ minutes. I wish to find a general formula for the population if one unit of time is $50$ minutes for $t=0,1,2,3,\ldots$.

If one unit of time is $20$ minutes, then I can 'easily' find $N(t)=12\cdot 2^t,~t=0,1,2,\ldots$.

How do I solve the original problem? Is there a general method for approaching these types of questions?

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If it takes 20 minutes = $1t$ to double, it takes $20/50t'$ to double, because $t' = 50/20t$. This gives $N(t) = 12 \cdot 2^{\frac {50t}{20}}$ –  Sam DeHority Jan 23 '13 at 20:19
    
Can I also say that there are two and half 20 minutes in one 50 minutes, so $N(t)=12\cdot 2^{2.5t}$? –  Gorg Jan 23 '13 at 20:27
    
Gorg Yes, that would work! There are 50 minutes equals 2.5 units of (20 minutes) –  amWhy Jan 23 '13 at 20:34
    
@amWhy: Thanks a bunch. –  Gorg Jan 23 '13 at 20:38
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2 Answers

up vote 2 down vote accepted

The "doubling time" becomes $\dfrac{20}{50} = \dfrac{2}{5}$ units, so $$N(t) = 12\cdot 2^{t/(2/5)} = 12 \cdot 2^{\Large\left(\frac{5t}{2}\right)} = 12\cdot2^{(2.5t)}$$

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The doubling time is now $20/50 = 0.4$ units, so $N(t) = 12 \cdot 2^{t/0.4}$, with $t$ in units of time.

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