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Chapter 12 Exponential functions

Section 12.1 Basics

Example 12.1.1. A doubling rate of growth.

An environmental scientist is studying the rate of population growth of a species of tree in a geographic area by making yearly observations of new seedlings and newly fallen trees. She observes that the rate of growth is increasingly dramatic, approximately doubling every year.
Data points for rate of growth of a population.
Figure 12.1.2. Data points for rate of growth of a population. Note that axes are not the same scale.
To create a mathematical model from the data points, the researcher fits a curve to them.
Fitted curve for rate of growth of a population.
Figure 12.1.3. Fitted curve for rate of growth of a population.
If we assign a scale on both axes in Figure 12.1.3 so that the grid lines fall at integer values, the the plotted points fall at
\begin{align*} r(0) \amp = 1 \amp r(1) \amp = 2 \amp r(2) \amp = 4 \amp r(3) \amp = 8 \amp r(4) \amp = 16\text{,} \end{align*}
as the height of each data point is double the height of the preceding point, in line with the “doubling every year” observation of the researcher, and we can model these data points with the formula
\begin{equation*} r(t) = 2^t \text{.} \end{equation*}
If we plot some additional points, we find the fitted curve matches other values of this rate function.
Plotting additional rate function values against our "doubling" model curve.
Figure 12.1.4. Plotting function values \(r(-1) = 2^{-1} \) and \(r(1/2) = 2^{1/2} \) on our “doubling” model curve.
It appears that the model curve matches the values of the function \(r(t) = 2^t \) not only at whole, positive exponents but also at
  • negative exponents, which are used to mean reciprocal:
    \begin{equation*} 2^{-a} = \frac{1}{2^a} \end{equation*}
  • rational exponents, which are used to combine a power with a root:
    \begin{equation*} 2^{a/b} = \sqrt[b]{2^a} \text{.} \end{equation*}
We are not yet ready to attach a meaning to irrational exponents, such as
\begin{equation*} 2^{\pi} \text{,} \end{equation*}
but for the moment we will assume that our formula \(r(t) = 2^t \) matches our model curve at all values of \(t \text{.}\)

Definition 12.1.5. Exponential function with base \(a \).

For positive constant \(a \text{,}\) the exponential function with base \(a \) is the function
\begin{equation*} f(t) = c a^t \text{,} \end{equation*}
where the constant \(c \) represents the initial value \(c = f(0) \text{.}\)
As the constant \(c \) merely represents a vertical scaling and/or reflection (if \(c \) is negative), we will focus our further investigation of exponential functions on the case \(c = 1 \text{.}\)
Exponential functions come in three varieties:
  • exponential growth when \(a \gt 1 \)
  • constant when \(a = 1 \)
  • exponential decay when \(0 \lt a \lt 1 \text{.}\)
A typical exponential growth graph.
(a) A typical exponential growth graph.
A typical exponential decay graph.
(b) A typical exponential decay graph.
Figure 12.1.6. Graphs of exponential functions.

Section 12.2 Basic properties of exponential functions

From the graphs in Figure 12.1.6 we have some obvious properties.
The graphs of all exponential functions look essentially the same because they are all horizontal transformations of each other.

Justification.

Since the “new” base \(b \) is a positive number, it is in the range of the exponential function \(a^t \text{.}\)
New base as an output of an exponential function.
Figure 12.2.4. New base as an output of an exponential function.
We can use Pattern 11.2.10 to determine the value of \(t \) so that \(a^t = b \text{:}\)
\begin{gather*} a^t = b \\ \ln(a^t) = \ln(b) \\ t \ln(a) = \ln(b) \\ t = \frac{\ln(b)}{\ln(a)} \text{.} \end{gather*}
Substituting \(b = a^{\bbrac{\ln(b) / \ln(a)}} \) into the exponential function formula \(b^t \text{,}\) we obtain
\begin{equation*} b^t = {(a^{\bbrac{\ln(b) / \ln(a)}})}^t = a^{t \bbrac{\ln(b) / \ln(a)}} \text{.} \end{equation*}
Graph of one exponential function as a horizontal scaling of another.
Figure 12.2.5. Graph of one exponential function as a horizontal scaling of another.

Remark 12.2.6.

The transformation in Pattern 12.2.3 will involve a horizontal reflection precisely when one of the bases is greater than one and the other is less than one. That is, when one exponential function is a growth function and the other is a decay function.

Section 12.3 The natural exponential function

Subsection 12.3.1 Euler’s number

In Pattern 12.2.3, if we choose \(a \) so that \(\ln(a) = 1 \text{,}\) then the comparison between the two exponential formulas becomes much simpler:
\begin{equation} b^t = a^{t \ln(b)} \text{.}\tag{✶} \end{equation}
Just like we use \(\pi \) to represent the special number that corresponds to the area of a unit circle, the special number so that \(\ln(a) = 1 \) is given its own symbol.

Definition 12.3.1. Euler’s number.

The real number, denoted by the letter \(e \text{,}\) so that \(\ln(e) = 1 \text{.}\)
Biographical Aside: Leonhard Euler.
Euler’s number defined relative to a special rate function.
Figure 12.3.2. Euler’s number defined relative to the special rate function \(r(t) = 1/t \text{.}\)
Sage knows about Euler’s number e. We can express it as a (truncated) decimal number using the n() method.
As \(a = e \) is the special value that makes (✶) work, it becomes “natural” to compare every exponential function to the exponential function with base \(e \text{.}\)

Definition 12.3.3. Natural exponential function.

The function \(\exp(t) = e^t \text{,}\) where the constant base \(e \) is Euler’s number.

Subsection 12.3.2 The natural exponential function as a reflection of the natural logarithm

It is possible to define the natural exponential function without using exponential notation — this will help us to make sense of how to compute an exponential function’s output for irrational inputs. Consider what happens when we reflect the graph of the natural logarithm in the line \(y = t \text{.}\)
A reflection of the graph of the natural logarithm function.
Figure 12.3.5. A reflection of the graph of the natural logarithm function.
This reflected curve passes the Vertical Line Test, and so it is the graph of some function. Indeed, it looks like the graph of an exponential function, but which one? Let’s explore by plotting some points. A graph \(y = f(t) \) is a collection of input-output pairs \((t,y) \text{.}\) Reflecting in the line \(y = t \) interchanges the coordinates in a point, so that if \((t,y) \) is on the graph \(y = \ln(t) \text{,}\) then \((y,t) \) is on the reflected curve. For example, from Pattern 11.2.1, we have
\begin{gather*} \ln(1) = 0 \\ \implies (1,0) \text{ on graph } y = \ln(t) \\ \implies (0,1) \text{ on reflected curve } \text{.} \end{gather*}
Now, the point \((0,1) \) is on the graph of every exponential function, as \(a^0 = 1 \) for each positive base \(a \text{.}\) So let’s consider some other points. Using Definition 12.3.1 along with Pattern 11.2.10, we have
\begin{equation*} \ln(e^r) = r \ln(e) = r \cdot 1 \text{,} \end{equation*}
so
\begin{gather*} \ln(e^r) = r \\ \implies (e^r,r) \text{ on graph } y = \ln(t) \\ \implies (r,e^r) \text{ on reflected curve } \text{.} \end{gather*}
Specific points on a reflection of the graph of the natural logarithm function.
Figure 12.3.6. Specific points on a reflection of the graph of the natural logarithm function.
Since \((r,e^r) \) is on the reflected curve for all values of \(r \text{,}\) we can say that this is indeed the graph of the natural exponential function.

Remark 12.3.8.

We will further explore pairs of functions whose graphs are reflections of each other in the line \(y = t \text{,}\) similarly to the natural exponential and logarithm functions, in Chapter 17.

Subsection 12.3.3 Computing values of exponential functions

Realizing the natural exponential function as a reflection of the natural logarithm function, as in Pattern 12.3.7, tells us (in principle) how to compute expressions like \(e^\pi \text{.}\)
Pattern 12.3.9 says that computing \(e^\pi \) is equivalent to solving \(\ln(z) = \pi \) for \(z \text{.}\) In the Sage cell below, change the value of the z variable (which is used as the upper bound of the integration command) and re-evaluate until you get reasonably close to the value of \(\pi \text{.}\) (Recall that \(\ln(t) \) is defined as an integral of rate function \(r(t) = 1/t \text{.}\)) The first command in the Sage cell below prints out a numerical approximation of \(\pi \text{.}\)
Of course, computers and calculators know how to (approximately) compute values of \(e^t \) for arbitrary values of \(t \text{.}\)
We can combine Pattern 12.3.9 with Pattern 12.3.4 and use a similar method to compute values like
\begin{equation*} 2^\pi = e^{\pi \ln(2)} \end{equation*}
by solving
\begin{equation*} \ln(z) = \pi \ln(2) \text{.} \end{equation*}
Use the Sage cell below to do just that. As before, you are attempting to refine the value of the z variable until the integration result is very close to \(\pi \ln(2) \text{,}\) which the first command computes for you.
When you’ve had enough of trial-and-error attempts to determine the value of \(2^\pi \) above, you can ask Sage to compute it for you below.

Subsection 12.3.4 Undoing the natural exponential and logarithm functions

The natural exponential and logarithm functions are not only graphical reflections of each other, they “reverse” each other algebraically, just as cube and cube root reverse each other.

Aside: Warning.

We can use the patterns above to solve equations involving \(\exp \) and \(\ln \text{.}\) Before we do examples, we continue the analogy with solving equations involving cube and cube root. Suppose you wanted to solve the equation
\begin{equation*} \sqrt[3]{t + 1} = 2 \text{.} \end{equation*}
The first step is to “undo” the cube root operation by cubing, to free the variable \(t \) from being trapped inside the cube root operation. But to maintain the equality, we must cube both sides:
\begin{gather*} {(\sqrt[3]{t + 1})}^3 = 2^3 \\ t + 1 = 8 \\ t = 7 \text{.} \end{gather*}
Similarly, to solve the equation
\begin{equation*} {(t + 1)}^3 = 2 \text{,} \end{equation*}
the first step is to “undo” the cube operation on the left by taking the cube root of both sides:
\begin{gather*} \sqrt[3]{{(t + 1})^3} = \sqrt[3]{2} \\ t + 1 = \sqrt[3]{2} \\ t = \sqrt[3]{2} - 1 \text{.} \end{gather*}
We can perform similar algebra using \(\exp \) and \(\ln \text{.}\)

Example 12.3.11. Solving an equation involving the natural exponential function.

To solve the equation
\begin{equation*} e^{t + 1} = 2 \text{,} \end{equation*}
we first “undo” the exponential by taking the logarithm of both sides and applying Property 1 of Pattern 12.3.10:
\begin{gather*} \ln(e^{t + 1}) = \ln(2) \\ t + 1 = \ln(2) \\ t = \ln(2) - 1 \text{.} \end{gather*}
Since every exponential is really just a horizontal transformation of the natural one, we can use the same technique to solve an equation involving any exponential.

Example 12.3.12. Solving an equation involving an exponential function.

To solve the equation
\begin{equation*} a^{t + 1} = 2 \text{,} \end{equation*}
we first take the logarithm of both sides and apply Pattern 11.2.10:
\begin{gather*} \ln(a^{t + 1}) = \ln(2) \\ (t + 1) \ln(a) = \ln(2) \\ t + 1 = \frac{\ln(2)}{\ln(a)} \\ t = \frac{\ln(2)}{\ln(a)} - 1 \text{.} \end{gather*}
Finally, an example with the roles of \(\exp \) and \(\ln \) reversed.

Example 12.3.13. Solving an equation involving the natural logarithm function.

To solve the equation
\begin{equation*} \ln(t + 1) = 2 \text{,} \end{equation*}
we first “undo” the logarithm by taking the natural exponential of both sides and applying Property 2 of Pattern 12.3.10:
\begin{gather*} e^{\ln(t + 1)} = e^2 \\ t + 1 = e^2 \\ t = e^2 - 1 \text{.} \end{gather*}

Section 12.4 Logarithms in other bases

Every exponential function is a horizontal scaling (possibly including a reflection) of the natural exponential function. (See Pattern 12.3.4.) In the reflection in the line \(y = t \) between the graphs of \(\exp(t) \) and \(\ln(t) \text{,}\) a horizontal scaling of \(\exp(t) \) corresponds to a vertical scaling of \(\ln(t) \text{.}\)
A reflection of the graph of a horizontal scaling of the natural exponential function.
Figure 12.4.1. A reflection of the graph of a horizontal scaling of the natural exponential function.
We know that
\begin{equation*} a^t = \exp\bbrac{t \ln(a)}\text{,} \end{equation*}
a horizontal scaling by scale factor \(\ln(a) \text{.}\) In the case that \(a \gt e \) (so that \(\ln(a) \gt 1 \)), the scaling will be a compression, as pictured in Figure 12.4.1. This corresponds to a vertical compression of the graph of \(\ln(t) \text{,}\) by the same scale factor. But to achieve a vertical compression, we must divide instead of multiply. Therefore, the scaled version of the natural logarithm pictured in Figure 12.4.1 is
\begin{equation*} \ell(t) = \frac{\ln(t)}{\ln(a)} \text{.} \end{equation*}

Definition 12.4.2. Base-\(a \) logarithm.

For \(a \gt 0 \text{,}\) define the base-\(a \) logarithm to be the function
\begin{equation*} \log_a(t) = \frac{\ln(t)}{\ln(a)} \text{.} \end{equation*}

Note 12.4.3.

Since \(\ln(e) = 1 \text{,}\) a logarithm in base \(a = e \) is just the natural logarithm function \(\ln(t) \text{.}\)
Every logarithm function enjoys the same properties as the natural logarithm described in Section 11.2. And the relationship between logarithm and exponential of the same base also carries over.
In particular, note that \(\log_a (a) = 1 \text{.}\)

Section 12.5 More properties of exponential functions

Now that we have established the “reflection” relationship between the natural exponential and logarithm functions, we can “reflect” the properties from Section 11.2 to obtain further properties of exponential functions.

Justification.

From Pattern 12.3.4, we have
\begin{equation*} a^{u + v} = e^{(u + v) \bbrac{\ln(a)}} \text{.} \end{equation*}
From Pattern 12.3.9, \(z = e^{(u + v) \bbrac{\ln(a)}} \) should be the unique value so that \(\ln(z) = (u + v) \bbrac{\ln(a)} \text{.}\) Let’s check whether applying the natural logarithm to \(a^u a^v \) evaluates to the same result. Using the properties of the natural logarithm, we have
\begin{align*} \ln(a^u a^v) \amp = \ln(a^u) + \ln(a^v) \\ \amp = u \ln(a) + v \ln(a) \\ \amp = (u + v) \ln(a) \text{,} \end{align*}
as desired.

Remark 12.5.2.

Compare patterns:
  • Pattern 12.5.1.
    The exponential of a sum is the product of the exponentials.
  • Pattern 11.2.4.
    The logarithm of a product is the sum of the logarithms.
The “reflection” here is the interchanging of the words sum and product. Try to identify the “reflection” in each of the next two properties, compared to the corresponding property in Section 11.2. Also try to provide a justification for each of these two properties in the same manner as the justification for Pattern 12.3.4.
Here is another algebraic property that doesn’t have a corresponding “reflected” property in Section 11.2.

Justification.

From Pattern 12.3.4, we have
\begin{align*} {(a^u)}^v \amp = e^{v \ln(a^u)} \amp a^{u v} \amp = e^{(u v) \ln(a)} \text{.} \end{align*}
But from Pattern 11.2.10 we have
\begin{equation*} \ln(a^u) = u \ln(a) \text{,} \end{equation*}
so the two exponential expressions above are indeed the same.

Note 12.5.6.

Pattern 12.2.2 could also be considered a “reflection” of the corresponding logarithmic pattern Pattern 11.2.3. When we reflect in the line \(y = t \text{,}\) we reflect the horizontal axis onto the vertical axis, and vice versa. So inputs become outputs and outputs become inputs. If we view the pattern
\begin{equation*} t \to 0^+ \qquad \implies \qquad \ln(t) \to -\infty \end{equation*}
from Pattern 11.2.3 as
\begin{equation*} \text{input} \to 0^+ \qquad \text{then} \qquad \text{output} \to -\infty \text{,} \end{equation*}
then the “reflected” pattern for the exponential function is
\begin{equation*} \text{output} \to 0^+ \qquad \text{when} \qquad \text{input} \to -\infty \text{,} \end{equation*}
so that
\begin{equation*} t \to -\infty \qquad \implies \qquad \exp(t) \to 0^+ \text{,} \end{equation*}
meaning that the exponential graph heads towards \(0 \) from above as the input goes off to the left.

Section 12.6 Derivatives of exponential functions

Subsection 12.6.1 Derivative of the natural exponential function

Let’s return to our visualization of the graph of the natural exponential function as a reflection of the graph of the natural logarithm function. Additionally, we have drawn on some tangent lines, one to each curve.
The graphs of the natural exponential and logarithm functions, with tangent lines.
(a) A tangent line to the natural logarithm function.
The graphs of the natural exponential and logarithm functions, with tangent lines.
(b) A tangent line to the natural exponential function.
Figure 12.6.1. The natural exponential function as a reflection of the graph of the natural logarithm function, with corresponding tangent lines on each curve.
The rise-over-run formula can be used between any two points on a line to compute its slope. For the tangent line to the natural logarithm function in Figure 12.6.1(a), we have arbitrarily chosen two points on the tangent line and labelled the run between them as \(\change{t} \) and the rise between them as \(\change{y} \text{,}\) so that the line’s slope is \(\inlineslope{y}{t} \text{.}\) The slope of this tangent line also has a relationship to the input point \(t^\ast \) at which the tangent line meets the curve, given by Pattern 11.3.1. So we can say that
\begin{equation*} \slope{y}{t} = \frac{1}{t^\ast} \text{.} \end{equation*}
By symmetry, in Figure 12.6.1(b) the output point \(y^\ast \) should have exactly the same relationship to \(\change{t} \) and \(\change{y} \text{,}\) under the following correspondences through the reflection \(t \leftrightarrow y \text{.}\) (In these correspondences, the left-hand side is from Figure 12.6.1(a) and the right-hand side is from Figure 12.6.1(b).)
\begin{align*} t^\ast \amp \leftrightarrow y^\ast \amp \change{t} \amp \leftrightarrow \change{y} \amp \change{y} \amp \leftrightarrow \change{t} \end{align*}
So we have
\begin{equation*} \slope{t}{y} = \frac{1}{y^\ast} \text{.} \end{equation*}
But now this slope is run-over-rise instead of rise-over-run; to get the correct slope for the tangent line in Figure 12.6.1(b), we need to flip things:
\begin{equation*} \slope{y}{t} = y^\ast \text{.} \end{equation*}
Since \(y^\ast \) is an output for the natural exponential function, we obtain the following derivative formula for the natural exponential function.
This is a remarkable fact! It says that a quantity that experiences exponential growth will also have a rate of growth that grows exponentially. This is why exponential growth can be both exciting (when considering compounding of gains in your investment account, for example) and terrifying (when considering unchecked spread of a virus in a population, for example): the more there is, the sooner there will be a lot more than that.

Example 12.6.3. Derivative involving an exponential.

For
\begin{equation*} q(t) = t^5 + 6 e^t - 24 \text{,} \end{equation*}
we have
\begin{align*} q'(t) \amp = \opddt \bbrac{t^5 + 6 e^t - 24} \\ \amp = \ddt{t^5} + \ddt{(6 e^t)} - \ddt{24} \\ \amp = 5 t^4 + 6 \ddt{e^t} + 0 \\ \amp = 5 t^4 + 6 e^t \end{align*}
where in the first step we have applied the Sum Rule and the Difference Rule, in the second step we have applied the Power Rule, the Constant Multiple Rule, and the rule for the derivative of a constant, and in the final step we have applied our new derivative formula for the natural exponential.

Subsection 12.6.2 Exponential functions as solutions to proportional differential equations

We can now combine Pattern 12.6.2 with Pattern 6.3.17 to verify exponential solutions to differential equations with form similar to our Radioactive decay example (Example 7.2.4), where the rate is proportional to the quantity.

Example 12.6.4. Verifying an exponential solution to a proportional differential equation.

Suppose \(q(t) \) is a quantity function whose rate of accumulation is proportional to the quantity, with proportionality constant \(k \text{.}\) For the purposes of this example, let’s take our proportionality constant to be \(k = 1 / 3 \text{,}\) so that
\begin{equation} \dqdt = \frac{ q(t) }{ 3 } \text{.}\tag{✶} \end{equation}
So we are looking for a function whose derivative is essentially the same function, but vertically compressed by a factor of \(3 \text{.}\) The pattern
\begin{equation*} \ddt{e^t} = e^t \end{equation*}
is almost correct, but here the vertical scale factor between function and derivative is only \(1 \text{.}\) However, we know from Pattern 6.3.17 that introducing a horizontal scale factor into the quantity function also introduces a vertical scale factor into the derivative function. In this case, if we take
\begin{equation*} q(t) = \exp(t / 3) \text{,} \end{equation*}
then Pattern 6.3.17 says that
\begin{align*} q'(t) \amp = \frac{ \exp'(t / 3) }{ 3 } \\ \amp = \frac{ \exp(t / 3) }{ 3 } \\ \amp = \frac{ q(t) }{ 3 } \text{.} \end{align*}
This shows that \(q(t) = e^{t / 3} \) is a solution to the differential equation in (✶).
This is only one particular solution to (✶). Can we guess at the general solution by introducing some parameter? Our particular solution above has initial value
\begin{equation*} q(0) = e^{0 / 3} = 1 \text{.} \end{equation*}
But exponential growth or decay does not turn into some other behaviour if there is a different initial value — otherwise a horizontal shift (so that the point in time that is considered \(t = 0 \) is shifted) would change the type of graph, which is not possible. For example, see our direction field for \(k = -1/2 \) (Figure 7.3.4) — the example solution curves all look like exponential decay no matter the initial value.
Recall from Definition 12.1.5 that introducing a constant multiple in an exponential function changes the initial amount. So let’s try
\begin{equation*} q(t) = A e^{t / 3} \text{,} \end{equation*}
where \(A \) represents the initial amount \(A = q(0) \text{.}\) Using the Constant Multiple Rule (Pattern 6.3.4) for our first step, we have
\begin{align*} q'(t) \amp = \opddt \bbrac{A \exp(t / 3)} \\ \amp = A \opddt \bbrac{\exp(t / 3)} \\ \amp = A \left( \frac{ \exp'(t / 3) }{ 3 } \right) \\ \amp = \frac{ A \exp(t / 3) }{ 3 } \\ \amp = \frac{ q(t) }{ 3 } \text{.} \end{align*}
This verifies that for every possible initial value \(A \text{,}\) the exponential function \(q(t) = A e^{t / 3} \) is a solution to the rate equation in (✶).
Our analysis at the end of Example 12.6.4 doesn’t verify that every solution has that form, but it strongly suggests it: given any initial amount \(A = q(0) \) we have a solution function \(q(t) = A e^{k t} \) that fits the behaviour of the differential equation, and it doesn’t seem possible for there to exist some other solution function \(q(t) \) with the same \(q(0) \) value, as that would mean the system would somehow have to initially “decide” which of those two solution curves to follow. For now we will at least record the general pattern that we’ve found without overstating our current knowledge, but in Chapter 16 we will be able to verify that we in fact have the general solution.

Remark 12.6.6. Calculations versus reality.

In the discussion between Example 12.6.4 and Pattern 12.6.5, we asserted the plausible-sounding principle that a system should always behave exactly the same way given the exact same initial conditions. In reality, unless you are counting molecules, it is nearly impossible to re-run an experiment with exactly the same initial conditions. And we will find in Chapter 21 that sometimes a particular initial quantity value \(A = q(0) \) can be unstable, in the sense that the system behaves very differently depending on whether the actual \(q(0) \) value is slightly greater than \(A \) or slightly less then \(A \text{.}\)

Example 12.6.7. Radioactive decay.

Suppose \(q(t) \) measures how many kilograms of a 3 kg hunk of radioactive material remain after \(t \) years. For this particular material, it is experimentally determined that its rate of decay can be modelled as
\begin{equation*} \dqdt = - \frac{q(t)}{2} \text{.} \end{equation*}
This is the same differential equation we analysed graphically in Example 7.3.2, and from the direction field produced there we expect solutions to have the “shape” of exponential decay. Based on our analysis in Example 12.6.4, we know that quantity function \(q(t) = 3 e^{- t / 2} \) is a solution to this differential equation, where we have used the initial amount \(q(0) = 3 \) as the value of parameter \(A \text{,}\) and the proportionality constant \(k = - 1 /2 \) from the differential equation.
We can use the solution function to determine properties of the material, such as its half-life, which is the time it takes for the sample to be reduced by half. Our initial quantity was \(q(0) = 3 \text{,}\) so we are looking to solve for the \(t \)-value that satisfies
\begin{equation*} q(t) = 3 / 2 = 1.5 \text{.} \end{equation*}
As in Subsection 12.3.4, we can use the natural logarithm to help us solve this.
\begin{gather*} 3 e^{- t / 2} = \frac{3}{2} \\ e^{- t / 2} = \frac{1}{2} \\ \ln(e^{- t / 2}) = \ln\left(\frac{1}{2}\right) \\ - \frac{t}{2} = - \ln(2) \\ t = 2 \ln(2) \end{gather*}
So it will take approximately \(1.4 \) years for the sample to be reduced by half.

Remark 12.6.8. Half-life is independent of initial amount.

For differential equation
\begin{equation*} \dqdt = -k q(t) \end{equation*}
(where here we have explicitly included a negative sign and assume that the constant \(k \) is positive, so that overall the constant of proportionality is negative and we have decay), the half-life depends only on \(k \) and not on the initial amount \(A = q(0) \text{.}\) That is, it will take the same amount of time for a sample to be reduced by half, regardless of how large or small of a sample. This is because we are looking to solve
\begin{equation*} A e^{- k t} = \frac{A}{2} \qquad \implies e^{- k t} = \frac{1}{2} \text{,} \end{equation*}
where the value of \(A \) is no longer involved after one step of simplification.

Checkpoint 12.6.9. Half-life formula.

Determine a formula in \(k \) for the half-life of a radioactive material that decays according to
\begin{equation*} \dqdt = -k q(t) \text{,} \qquad k \gt 0 \text{.} \end{equation*}

Subsection 12.6.3 Derivatives of general exponential functions

Since every exponential function is a horizontal scaling of the natural exponential function (Pattern 12.3.4), we may combine our derivative formula for the natural exponential function (Pattern 12.6.2) with our rule for the derivative of a horizontally-scaled function (Pattern 6.3.17).

Justification.

Consider \(q(t) = a^t \) with \(a \gt 0 \text{.}\) The formula we would like to verify is
\begin{equation*} q'(t) = q(t) \cdot k \quad \text{for} \quad k = \ln(a) \text{.} \end{equation*}
With this same value of \(k \text{,}\) Pattern 12.3.4 says that
\begin{equation*} q(t) = \exp(k t) \text{.} \end{equation*}
Applying Pattern 6.3.17 with \(\exp \) in place of \(p \) gives us
\begin{equation*} q'(t) = k \exp'(k t) \text{.} \end{equation*}
But \(\exp'(t) = \exp(t) \text{,}\) so
\begin{equation*} q'(t) = k \exp(k t) = k q(t) \text{,} \end{equation*}
our desired result.

Section 12.7 Integrals of exponential functions

Suppose we have an exponential rate function, \(r(t) = e^t \text{.}\) This means that the rate of accumulation of some quantity is itself growing exponentially, similar to the rate doubles every year pattern in Example 12.1.1. We would like to know the associated accumulation function based at \(t = 0 \text{,}\)
\begin{equation*} A_0(t) = \ccmint{0}{t}{e^u}{u} \text{.} \end{equation*}
For power functions we have found that the pattern of integrate-then-differentiate returns us back to the original function. Here we remind ourselves of that pattern, in general for power functions with positive exponent \(m \gt 0 \text{,}\) and also specifically for the power function with exponent \(m = -1 \) whose integral defines the natural logarithm function.
\begin{align*} \opddt \, \ccmint{0}{t}{ u^m }{u} \amp = t^m \amp \opddt \, \ccmint{0}{t}{ \frac{1}{u} }{u} \amp = \frac{1}{t} \end{align*}
If we expect this pattern to also hold for exponential functions, then we should have
\begin{equation*} \opddt \ccmint{0}{t}{ e^u }{u} = e^t \text{,} \end{equation*}
or
\begin{equation*} A_0(t) = \ccmint{0}{t}{ e^u }{u} \qquad \implies \qquad \ddt{A_0} = e^t \text{.} \end{equation*}
What kind of function has an exponential derivative? Well, the natural exponential function does! (Pattern 12.6.2) But we have to be careful: just as recording only speedometer data will not tell you where the vehicle ended up at the end of the trip, knowing only a derivative calculation result does not tell you the original function exactly. You also need to take into account the initial condition. So while we have both
\begin{align*} \ddt{A_0} \amp = e^t \text{,} \amp \ddt{(\exp)} \amp = e^t \text{,} \end{align*}
we find a mismatch in their initial values:
\begin{align*} A_0(0) \amp = \ccmint{0}{0}{ e^ u }{u} = 0 \text{,} \amp \exp(0) \amp = e^0 = 1 \text{.} \end{align*}
So it seems that, similar to Formula 2 in Pattern 10.4.10, we need a “correction term” in our formula for \(A_0(t) \text{,}\) where the correction amount is precisely the amount of mismatch in initial value that we found above:
\begin{equation*} \ccmint{0}{t}{ e^u }{u} = e^t - 1 \text{.} \end{equation*}
In the case that the exponential rate function has a different base, we also need a vertical scale correction factor. (Compare with the vertical scale factor in the derivative formula of a general exponential function in Pattern 12.6.10.)
Graph demonstrating that the area under the natural exponential function is measured by the natural exponential function.
Figure 12.7.2. Area under the natural exponential function is measured by the natural exponential function.

Example 12.7.3. Using Pattern 12.7.1.

Let’s extend Example 11.1.4 by computing
\begin{equation*} \ccmint{3}{4}{ t + \frac{1}{t} + e^t }{t} \text{.} \end{equation*}
We’ll use the same strategy as in Example 11.1.4 and Example 10.4.14, but here since we only have a formula for the exponential integral based at initial time \(t_0 = 0 \text{,}\) we’ll first split into two integrals.
\begin{align*} \amp \ccmint{3}{4}{t + \frac{1}{t}}{t} \amp \amp \ccmint{3}{4}{ e^t }{t}\\ \amp = \ccmint{1}{4}{t + \frac{1}{t}}{t} - \ccmint{1}{3}{t + \frac{1}{t}}{t} \amp \amp = \ccmint{0}{4}{e^t}{t} - \ccmint{0}{3}{e^t}{t}\\ \amp = \left( \frac{4^2 - 1}{2} + \ln(4) \right) - \left( \frac{3^2 - 1}{2} + \ln(3) \right) \amp \amp = (e^4 - 1) - (e^3 - 1)\\ \amp = \frac{7}{2} + \ln\left(\frac{4}{3}\right) \amp \amp = e^4 - e^3 \end{align*}
Our original integral is now just the sum of these two integrals.
\begin{align*} \ccmint{3}{4}{ t + \frac{1}{t} + e^t }{t} \amp = \ccmint{3}{4}{t + \frac{1}{t}}{t} + \ccmint{3}{4}{ e^t }{t}\\ \amp = \frac{7}{2} + \ln\left(\frac{4}{3}\right) + e^4 - e^3 \end{align*}
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