In a certain culture of bacteria, the number of bacteria increased sixfold in 10h. Assuming natural growth, how long did it take for their number to double?

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- Feb 25th 2010, 10:56 AMAbaloHelp with calculus problem
In a certain culture of bacteria, the number of bacteria increased sixfold in 10h. Assuming natural growth, how long did it take for their number to double?

- Feb 25th 2010, 11:05 AMe^(i*pi)
This is classic exponential growth so use the equation A(t) = A_0e^{kt}

where:

- $\displaystyle A(t)$ = Amount at time
*t* - $\displaystyle A_0$ = Amount at time 0/initial amount
- $\displaystyle k$ = growth constant
- $\displaystyle t$ = time

By definition the amount at time 0 is equal to $\displaystyle A_0$

If the increase is sixfold after ten hours we get the following:

$\displaystyle 6A_0 = A_0e^{10k}$ which is equal to $\displaystyle 6 = e^{10k}$

From there you can find k

Once you know your value of k find t using the rearranged equation below and that $\displaystyle A_2 =2A_0= A_0e^{kt}$

$\displaystyle t = \frac{1}{k} \left[ \ln (A_t) - \ln (A_0)\right]$ - $\displaystyle A(t)$ = Amount at time