Suppose $Q(t_1)=Q_1$ and $Q(t_2)=Q_2$ are the values of a variable $Q(t)$ at time $t_1$ and at a later time $t_2$. The growth factor per time interval $t_2-t_1$ equals:

$$g_{t_2 - t_1} = \frac{Q_2}{Q_1}$$ So the growth factor is $g$ per single unit of time is equal to: $$g_1 = g = \left(\frac{Q_2}{Q_1}\right)^{\frac{1}{t_2-t_1}}$$ You can now use these points to calculate the constant $c$, for example $$Q_1=c\cdot \left(\frac{Q_2}{Q_1}\right)^{\frac{t_1}{t_2-t_1}}$$ From this follows: $$c=Q_1\cdot \left(\frac{Q_1}{Q_2}\right)^{\frac{t_1}{t_2-t_1}}$$

Suppose that you took a chicken leg from the refrigerator at one o'clock in the afternoon and put it on a plate to reach room temperature.
Suppose room temperature was reached after two hours and the number of salmonella bacteria on the meat grows exponentially after this moment.
At six o'clock in the evening there are a hundred thousand bacteria and at nine o'clock in the evening, this number increased to 1.6 million.

Give the formula for the number of bacteria $A(t)$, in which $t$ is the time in hours, with $t=0$ at three o'clock in the afternoon.
Round the growth factor to four decimal numbers.

Explanation

$\displaystyle g_{\mathrm{3\,hours}}=\frac{1\,600\,000}{100\,000}=16$
$\displaystyle g_{\mathrm{hour}}=16^{\frac{1}{3}}=2^{\frac{4}{3}}\approx 2.5198$
This corresponds to a doubling time of 3/4 hours = 45 minutes.

$A(t)=c\cdot g^t$, $g=2.51984$ and $A(3)=100\,000$ leads to the equation $100000= c\cdot 2.51984^3$.
So $c=6250$.
We have found the following formula:

$$A(t)=6250\cdot 2.5198^t = 6250\cdot 2^{\frac{t}{0.75}}$$
Check: from the data you can see that the number of bacteria is multiplied by sixteen every three hours. Three hours before the clock strikes six, accidentally time $t=0$, the number of bacteria is 16 times less than the given number at six o'clock. So $A(0)= 100000/16=6250$.