# The count in a bacteria culture was 700 after 20 minutes and 1000 after 40 minutes. What was the initial size of the culture?

Jun 16, 2017

490 microorganisms.

#### Explanation:

I will assume exponential growth for bacteria. This means that we can model the growth with an exponential function:

$f \left(t\right) = {A}_{0} {e}^{k t}$

where $k$ is the growth constant and ${A}_{0}$ is the initial amount of bacteria.

Sub the two known values into the function to get two equations:

$700 = {A}_{0} {e}^{20 k}$ (1)

$1000 = {A}_{0} {e}^{40} k$ (2)

Divide (2) by (1) to find $k$:

$\frac{1000}{700} = \frac{\cancel{{A}_{0}} {e}^{40 k}}{\cancel{{A}_{0}} {e}^{20 k}}$

$\frac{10}{7} = {e}^{40 k - 20 k} = {e}^{20 k}$

Take the natural log of both sides to isolate $k$:

$\ln \left(\frac{10}{7}\right) = \cancel{\ln} {\cancel{e}}^{20 k}$

$\ln \left(\frac{10}{7}\right) = 20 k$

$k = \ln \frac{\frac{10}{7}}{20}$

Now that we have the growth constant, $k$, we can substitute one of the points in to solve for the initial amount, ${A}_{0}$:

$\left(40 , 1000\right)$

$1000 = {A}_{0} {e}^{\ln \frac{\frac{10}{7}}{20} \cdot 40}$

${A}_{0} = \frac{1000}{e} ^ \left(0.0178 \cdot 40\right) = 490$

Jun 16, 2017

Initial culture size was $490$

#### Explanation:

The growth can be considered as a geometric progression with the same rate of growth after each interval of $20$ minutes.

The rate of growth can be determined by $\frac{1000}{700} = \frac{10}{7}$

In terms of the size of the initial population $\left(x\right)$

This means:

$x \times \frac{10}{7} \rightarrow 700 \times \frac{10}{7} \rightarrow 1000$
$0 \text{ mins"color(white)(xxx) 20 " mins"color(white)(xxx) 40 " mins}$

So if we reverse the process we just divide by $\frac{10}{7}$

$x \leftarrow \frac{10}{7} \div 700 \leftarrow \frac{10}{7} \div \leftarrow 1000$

Remember that $\div \frac{10}{7} = \times \frac{7}{10}$

$1000 \times \frac{7}{10} = 700$

$700 \times \frac{7}{10} = 490$