EXP

Returns the mathematical constant e raised to the power of a number.

Syntax:

EXP(number)

returns $e^{n u mb er}$ .

Example:

EXP(1)

returns 2.71828182845904, the mathematical constant e to Zapof function's accuracy.

Application:

Bacterial Growth

Imagine you start with a single bacterium in a petri dish. You observe that this type of bacteria doubles every hour. We can model this growth using the exponential function.

Initial population (P0): 1 bacterium
Growth rate (k): We know the population doubles every hour. If $P (t) = P_{0} e^{k t}$ , and we know $P (1) = 2 P_{0}$ , then $2 P_{0} = P_{0} e^{k \cdot 1}$ . This simplifies to $2 = e^{k}$ , which means $k = ln (2) \approx 0.693$ .

So, our specific model for this bacterial colony's population is:

$P (t) = 1 \cdot e^{0.693 t}$

Here, the EXP function, $e^{0.693 t}$ , represents the factor by which the population has multiplied after t hours.

Table of Population Growth over Time

The table below shows the population of the bacteria colony at different time intervals, calculated using the EXP function.

	Time (t) in hours	Calculation ( $P (t) = e^{0.693 t}$ )	Population (P)
	A	B	C
1	0	$e^{0.693 \cdot 0} = e^{0}$	1
2	1	$e^{0.693 \cdot 1} \approx e^{0.693}$	2
3	2	$e^{0.693 \cdot 2} \approx e^{1.386}$	4
4	3	$e^{0.693 \cdot 3} \approx e^{2.079}$	8
5	4	$e^{0.693 \cdot 4} \approx e^{2.772}$	16
6	5	$e^{0.693 \cdot 5} \approx e^{3.465}$	32

Explanation:

At t=0, the exponent is 0, and e0=1. This is our initial population.
At t=1, the exponent is 0.693, and e0.693 is approximately 2. The population has doubled.
At t=2, the exponent is 2×0.693, and e1.386 is approximately 4. The population has doubled again from the previous hour, and is four times the initial population.
This pattern continues, with the population doubling every hour, demonstrating the powerful and rapid growth characteristic of the exponential function. The EXP function, in this case, is the engine that drives this compound growth over time.