One reason: Exponential trends show up a LOT in environmental science (the proportional change is the same over each time span)
Math reason: Turns out it’s a very useful value for calculus

EDS 212: Day 2, Lecture 1
The exponential function, logarithms, and limits
August 4th, 2025
Overview of Day 2 topics
The number \(e\)
The number \(e\) is a constant in mathematics, it is approximately \(e\approx 2.72\), but really its decimals continue infinitely without any pattern:
\(e = 2.71828182845904523536...\)
It’s exact value can be described as \[ e = 1 + 1 + \frac{1}{1*2} + \frac{1}{1*2*3} + \frac{1}{1*2*3*4} + \frac{1}{1*2*3*4*5} + ... \]
Like any other number, \(e\) follows normal power properties:
\[ \begin{align} e^x\cdot e^y &=& ?\\ \frac{1}{e^x} &=& ?\\ \frac{e^x}{e^y} &=& ?\\ (e^x)^r &=& ? \end{align} \]
✏️ Take a moment to write these with a single exponent.
The number \(e\)
The number \(e\) is a constant in mathematics, it is approximately \(e\approx 2.72\), but really its decimals continue infinitely without any pattern:
\(e = 2.71828182845904523536...\)
It’s exact value can be described as \[ e = 1 + 1 + \frac{1}{1*2} + \frac{1}{1*2*3} + \frac{1}{1*2*3*4} + \frac{1}{1*2*3*4*5} + ... \]
Like any other number, \(e\) follows normal power properties:
\[ \begin{align} e^{x+y}&=&e^x\cdot e^y \\ e^{-x}&=&\frac{1}{e^x}\\ e^{x-y}&=&\frac{e^x}{e^y}\\ e^{rx}&=&(e^x)^r \end{align} \]
The exponential funciton
The exponential function \(e^x\) is the number \(e\) raised to the \(x\) power.
In function notation you may sometimes see it as
\[ f(x) = e^x \ \ \text{ or } \ \ \text{exp}(x).\]
Why is the exponential function so common?
One reason: Exponential trends show up a LOT in environmental science (the proportional change is the same over each time span)
Math reason: Turns out it’s a very useful value for calculus
A real-world example:
Source: Our World in Data
But populations can’t grow exponentially forever
Gause, G. F. 1934. The Struggle for Existence. Baltimore: Williams and Wilkins.
Logistic growth
\[N_t=\frac{K}{1+[\frac{K-N_0}{N_0}]e^{-rt}}\]
Where \(N_t\) is the population size at time \(t\), \(K\) is the carrying capacity, \(N_0\) is the initial population size, and \(r\) is a growth rate.
✏️ What should be the value of the formula when \(t=0\)? Confirm your answer by substituting.
We should always understand the components of formulas and equations - both conceptually and mathematically.
Logistic growth
\[N_t=\frac{K}{1+[\frac{K-N_0}{N_0}]e^{-rt}}\]
Why might we expect logistic growth for many populations?
What variables besides time would influence the actual population?
Logarithms
\(\log_a(b)\) is the exponent you need to raise \(a\) to get a value of \(b\)
For example:
Natural logarithms
✏️
Natural logarithms
If \(y>0\), then \(\ln(x)\) is the exponent you need to raise \(e\) to in order to get \(y\).
This means:
\[ \begin{align} \ln(e^x) &= x \\ e^{\ln(y)} &= y \end{align} \]
In other words, \(\ln(x)\) is the inverse of the exponential function \(e^x\).
✏️ Examples
\(\ln(e^3)\)
\(\ln(\frac{1}{e})\)
\(\ln(1)\)
\(\ln(0)\)
\(\ln(-7)\)
Natural logarithms
If \(y>0\), then \(\ln(x)\) is the exponent you need to raise \(e\) to in order to get \(y\).
This means:
\[ \begin{align} \ln(e^x) &= x \\ e^{\ln(y)} &= y \end{align} \]
In other words, \(\ln(x)\) is the inverse of the exponential function \(e^x\).
Examples
\(\ln(e^3) = 3\) because 3 is the power of \(e\) needed to get \(e^3\).
\(\ln(\frac{1}{e}) = \ln(e^{-1}) = -1\)
\(\ln(1) = 0\) because 0 is the exponent we need to raise \(e\) to to get 1
\(\ln(0)\) does not exist, because there’s no number we can raise \(e\) to to get 0
\(\ln(-7)\) does not exist, because \(e\) to any number always be positive
Natural log properties

✏️
Natural log properties

Properties
\(\ln (e^x)=x\)
\(\ln(xy)=\ln x+\ln y\)
\(\ln(\frac{x}{y})=\ln x-\ln y\)
\(\ln(x^y)=y \ln x\)
Using logarithms to solve equations
✏️ Solve for \(t\) in the equation \(Pe^{rt} = A\).
Properties
\(\ln (e^x)=x\)
\(\ln(xy)=\ln x+\ln y\)
\(\ln(\frac{x}{y})=\ln x-\ln y\)
\(\ln(x^y)=y \ln x\)
Using logarithms to solve equations
✏️ Solve for \(t\) in the equation \(Pe^{rt} = A\).
\[ \begin{align} Pe^{rt} &= A \\ e^{rt} & = \frac{A}{P} \\ rt &= \ln\left(\frac{A}{P}\right) \\ t &= \frac{\ln(A)-\ln(P)}{r} \end{align} \]
Limits
Definition
We say a number \(L\) is the limit of a function \(f(x)\) as its variable \(x\) approaches a number \(c\), if the function’s output values \(f(x)\) approach \(L\) when \(x\) get closer to \(c\).
We write this symbolically as:
\[ \large \lim_{x\to c} f(x)=L \]
We read this as: “The limit of \(f(x)\) as \(x\) approaches \(c\) is \(L\).”
Limits example: numerical
What is the limit of \(f(x)=2x^2-4\) as \(x\) approaches 2?
We need to examine the values of \(f(x)\) as \(x\) gets closer to 2 from both sides.
From both directions it looks like \(f(x)\) converges to 4.
So, we say that “The limit of \(2x^2-4\) as \(x\) approaches 2 is 4”. We can write it like this:
\[ \lim_{x\to 2}(2x^2-4)=4 \]
Limits example: graph
What is the limit of \(f(x)=2x^2-4\) as \(x\) approaches 2?
From both directions it looks like \(f(x)\) converges to 4.
So, \(\lim_{x\to 2}(2x^2-4)=4\).
Do limits always exist?
The function \(|x|\) is the absolute value function. It is such that
\[ |x| = \begin{cases} x, & \text{ if } x\geq 0 \\ -x, & \text{ if } x <0 \end{cases} . \]
✏️ What happens to \(f(x)\) at \(x=2\)?
✏️ Try calculating: \[ \lim_{x\to 2}\frac{|x-2|}{x-2}. \]
Hint: Try to do numerical approximations or draw the graph of this function.
Do limits always exist?
✏️ What happens to \(f(x)\) at \(x=2\)?
This function is not defined at \(x=2\) since it would require us to divide by zero, which is undefined.
Try calculating \(\lim_{x\to 2}\frac{|x-2|}{x-2}\).
Let’s investigate:
Do limits always exist?
The function \(\frac{|x-2|}{x-2}\) approaches different values from either side at \(x=2\).
Therefore… the limit does not exist!
Lunch break!
image: Flaticon.com