# Assume that the number of bacteria follows an exponential growth model: . The count in the bacteria culture was 100 after 10 minutes and 1200 after 30 minutes. What was the initial size of the culture?

Feb 17, 2018

$\text{Approx. } 29$.

#### Explanation:

Let, $N$ denote the no. of bacteria after $t$ minutes.

Given that, $N$ follows an exponential growth model, we get,

 N=kb^t............(k,b" const.)"......(star).

To determine the consts. $k \mathmr{and} b$, let us utilise the conds. :

$\left(i\right) : t = 10 , N = 100 , \mathmr{and} , \left(i i\right) : t = 30 , N = 1200$.

$\left(\star\right) , \mathmr{and} \left(i\right) \Rightarrow 100 = k {b}^{10.} \ldots \ldots \ldots \left(\star 1\right)$, and,

$\left(\star\right) , \mathmr{and} \left(i i\right) \Rightarrow 1200 = k {b}^{30.} \ldots \ldots \ldots \left(\star 2\right)$.

$\therefore \left(\star 2\right) \div \left(\star 1\right) \Rightarrow 12 = {b}^{20} \Rightarrow b = {12}^{\frac{1}{20}}$.

$\text{Then, by } \left(\star 1\right) , k = \frac{100}{b} ^ 10 = \frac{100}{{12}^{\frac{1}{20}}} ^ 10 , \mathmr{and} ,$

$k = \frac{100}{12} ^ \left(\frac{1}{2}\right) = \frac{100}{\sqrt{4 \times 3}} = \frac{50}{\sqrt{3}} = \frac{1}{3} \cdot 50 \sqrt{3}$.

With these $k \mathmr{and} b$, we have,

$N = \frac{1}{3} \cdot 50 \sqrt{3} \cdot {12}^{\frac{t}{20}}$.

To, get the initial size of the culture, we plug in $t = 0$, & get,

$N = \frac{1}{3} \cdot 50 \sqrt{3} \approx \frac{1}{3} \left(50\right) \left(1.7321\right) \approx 28.87 = 29$.