Skip to main content
Logo image

Calculus Concepts and Modelling An introduction to the concepts of calculus

Jeremy Sylvestre

Chapter 8 Logarithmic growth

Section 8.1 The natural logarithm

Pattern 7.4.2 tells us how to calculate a quantity’s accumulation for any rate model that is a power of \(t\text{,}\) except for the special case of \(r(t) = t^{-1}\text{.}\) It turns out there is no simple formula for the accumulation for this special rate function, so we will have to resort to using integral notation to represent it. For convenience, we will also give this special accumulation function a name.
🔗

Definition 8.1.1. Natural logarithm.

The function that returns the accumulation for a quantity that is varying according to rate function \(r(t) = 1/t\text{,}\) relative to “initial” time \(t = 1\text{:}\)
\begin{equation*} \ln t = \ccmint{1}{t}{\frac{1}{u}}{u} \text{.} \end{equation*}
🔗
🔗
An area under the reciprocal function.
(a) An accumulation under \(r(t) = 1/t\text{.}\)
🔗
The corresponding value on the graph of the natural logarithm function.
(b) The corresponding accumulation value.
🔗
Figure 8.1.2. The natural logarithm as accumulation under \(r(t) = 1/t\text{.}\)
🔗
In Figure 8.1.2, the height of the logarithm function’s graph on the right at the input \(t\) corresponds to the magnitude of the shaded blue area under the graph of the rate function \(r(t) = 1/t\) on the left.
🔗
Notice that the graph of the logarithm function in Figure 8.1.2.(b) shows negative accumulation values for \(0 \lt t \lt 1\text{.}\) This is because of Property 2 in Convention 6.3.12 — if we allow time to “run backwards”, then we consider the quantity to be dissipating instead of accumulating, so that
\begin{equation*} \ln t = \ccmint{1}{t}{\frac{1}{u}}{u} \end{equation*}
for \(0 \lt t \lt 1\) is the negative of the accumulation that occurs over time interval \(t \le u \le 1 \text{.}\)
🔗
An area under the reciprocal function.
(a) A “reverse” accumulation under \(r(t) = 1/t\text{.}\)
🔗
The corresponding value on the graph of the natural logarithm function.
(b) The corresponding negative accumulation value.
🔗
Figure 8.1.3. The natural logarithm as accumulation under \(r(t) = 1/t\text{.}\)
🔗

Example 8.1.4. Using the natural logarithm to represent integral values.

Suppose we would like to calculate
\begin{equation*} \ccmint{3}{4}{ t + \frac{1}{t} }{t} \text{.} \end{equation*}
Similarly to Example 7.4.6, we split into two integrals, and then use Pattern 6.3.14 to re-base our integrals to match our formulas.
\begin{align*} \ccmint{3}{4}{ t + \frac{1}{t} }{t} \amp = \ccmint{3}{4}{t}{t} + \ccmint{3}{4}{\frac{1}{t}}{t}\\ \amp = \left( \ccmint{0}{4}{t}{t} - \ccmint{0}{3}{t}{t} \right) + \left( \ccmint{1}{4}{\frac{1}{t}}{t} - \ccmint{1}{3}{\frac{1}{t}}{t} \right)\\ \amp = \left( \frac{4^2}{2} - \frac{3^2}{2} \right) + ( \ln 4 - \ln 3 )\\ \amp = \frac{7}{2} + \ln \left(\frac{4}{3}\right) \end{align*}
🔗
🔗

Remark 8.1.5.

It might seem like “cheating” to write \(\ln 4\) as the value of
\begin{equation*} \ccmint{1}{4}{\frac{1}{t}}{t} \text{,} \end{equation*}
since Definition 8.1.1 says that to calculate \(\ln 4\) we would need to calculate that very integral (and similarly for \(\ln 3\)). But this is an acceptable way to represent the value of this integral, as there are several ways to numerically approximate values of the natural logarithm — using Riemann sums, for example. Your calculator even has a \(\ln\) button.
🔗
🔗
🔗

Section 8.2 Properties of the natural logarithm

First, our convention about integrating over a time domain of length \(0\) leads immediately to a special value of the natural logarithm.
🔗
Next, we note the domain of the natural logarithm.
🔗

Justification.

Since the graph of our rate function \(r(t) = 1/t\) used to define the natural logarithm is continuous on \(t \gt 0\text{,}\) that rate function is integrable on any (finite) subdomain of \(t \gt 0\text{.}\) However, given our “initial” point at \(t = 1\) in the integration defining the logarithm, any second domain endpoint that lies in \(t \le 0\) will cause the singularity of \(r(t)\) at \(t = 0\) to be included, resulting in an integral that cannot be computed.
🔗
🔗
From the graph of the logarithm function (see Figure 8.1.2.(b) or Figure 8.1.3.(b)), we can see obvious long-term and singular behaviour.
🔗
There are also a few algebraic patterns to the logarithm function that are not immediately obvious from its graph or from its definition as an accumulation function. The first involves the logarithm of a product of two numbers.
🔗

Justification.

By definition,
\begin{equation*} \ln (a b) = \ccmint{1}{a b}{\frac{1}{u}}{u} \text{.} \end{equation*}
Using Property 5 of Pattern 6.3.11, we may write
\begin{align*} \ccmint{1}{a b}{\frac{1}{u}}{u} \amp = \ccmint{1}{a}{\frac{1}{u}}{u} + \ccmint{a}{a b}{\frac{1}{u}}{u} \\ \amp = \ln a + \ccmint{a}{a b}{\frac{1}{u}}{u} \text{.} \end{align*}
So we already have the first part of our desired formula, \(\ln a\text{.}\) What about the second integral
\begin{gather} \ccmint{a}{a b}{\frac{1}{u}}{u} \text{?}\tag{✶} \end{gather}
How does it compare to
\begin{equation*} \ln b = \ccmint{1}{b}{\frac{1}{s}}{s} \text{?} \end{equation*}
(We have changed the integration variable to help keep things separate during our comparison of these two integrals.) It appears that the domain in integral (✶) has been stretched by a factor of \(a\) compared to the domain in the integral defining \(\ln b\text{.}\) Remember that an integral is a calculation of (oriented) area under a graph, added up “slice by slice” (see Figure 6.3.3). For each “slice” at position \(s\) in the integral for \(\ln b\text{,}\) we can create a corresponding “stretched slice” at position \(u = a s\text{,}\) where if \(1 \le s \le b\) then also \(a \le a s \le a b\text{.}\)
🔗
A representative slice of area in the integral defining a natural logarithm value.
(a) A representative slice of area in the integral defining \(\ln b\text{.}\)
🔗
A representative slice of area in an integral related to the natural logarithm.
(b) A representative slice of area in integral (✶).
🔗
Figure 8.2.5. A slice of the area defining \(\ln b\) transforms into a slice of the area corresponding to integral (✶).
🔗
Because we are stretching horizontally by a factor of \(a\text{,}\) the slice at position \(u\) on the second graph will also be wider by a factor of \(a\text{:}\)
\begin{equation*} du = a ds \text{.} \end{equation*}
But we are only transforming our domain values, we are not actually stretching our rate graph. So the slice at \(u\) has both a different width and a different height compared to the original slice at position \(s\text{,}\) since we have moved to a different place on the rate graph:
\begin{equation*} r(u) = \frac{1}{u} = \frac{1}{a s} = \frac{r(s)}{a} \text{.} \end{equation*}
However, when calculate the “area” of this “stretched slice”, we find it’s the same area as the slice at position \(s\) for the \(\ln b\) integral:
\begin{equation*} r(u) \intspace du = \left( \frac{r(s)}{a} \right) \left( a \intspace ds \right) = r(s) \intspace ds \text{.} \end{equation*}
So as we add up “slices” of area to compute the integral for \(\ln b\text{,}\) or we add up “slices” of area to compute integral (✶), we will be adding up the same values for those slices of area at corresponding positions, and so we will come out to the same total area:
\begin{equation*} \ccmint{a}{a b}{\frac{1}{u}}{u} = \ccmint{1}{b}{\frac{1}{s}}{s} = \ln b \text{,} \end{equation*}
and thus
\begin{equation*} \ln (a b) = \ccmint{1}{a}{\frac{1}{u}}{u} + \ccmint{a}{a b}{\frac{1}{u}}{u} = \ln a + \ln b \text{.} \end{equation*}
🔗
🔗

Remark 8.2.6.

We can be a little more precise in our justification above by considering actual Riemann sums of rectangular areas when we compare integral (✶) with the integral for \(\ln b\text{.}\) What our “corresponding slices” argument essentially shows is that for each Riemann sum approximating the integral for \(\ln b\text{,}\) we can create a Riemann sum approximating integral (✶) that computes to the same value. So the pattern of upper and lower sums squeezing down to the value for the integral for \(\ln b\) must be the same as the pattern of upper and lower sums squeezing down to the value of integral (✶).
🔗
🔗
So the logarithm turns a product into a sum. What does it do to a reciprocal?
🔗

Justification.

This time we will deal with an actual rectangular area in a typical Riemann sum, instead of just working with “slices”. For the integral
\begin{equation*} \ln a = \ccmint{1}{a}{\frac{1}{s}}{s} \text{,} \end{equation*}
consider a typical rectangle at sample point \(s\) in an upper sum approximation. Because of the shape of the graph of the rate function, this sample point will appear at the left endpoint of the base of the rectangular area. Mark the right endpoint as \(s'\text{,}\) so that
\begin{equation*} \change{s} = s' - s \end{equation*}
🔗
A representative rectangular area in a Riemann sum approximating the integral defining a natural logarithm value.
Figure 8.2.8. A representative rectangular area in a Riemann sum approximating \(\ln a\text{.}\)
🔗
By taking reciprocals, we can create a “mirror-image” rectangular base between \(u' = 1 / s'\) and \(u = 1 / s\text{.}\) To create a rectangle that represents a typical rectangle in an upper sum approximation of the integral
\begin{gather} \ccmint{1 / a}{1}{\frac{1}{u}}{u} \text{,}\tag{✶✶} \end{gather}
we’ll take the sample point to be at \(u'\text{.}\)
🔗
A representative rectangular area in a Riemann sum approximating an integral related to the natural logarithm.
Figure 8.2.9. A representative rectangular area in a Riemann sum approximating integral (✶✶).
🔗
To relate the areas of these two rectangles, first calculate
\begin{align*} \change{u} \amp = u - u' \\ \amp = \frac{1}{s} - \frac{1}{s'} \\ \amp = \frac{s' - s}{s s'} \\ \amp = \frac{1}{s s'} \, \change{s} \text{.} \end{align*}
So the area of the rectangle in Figure 8.2.9 is
\begin{align*} r(u') \, \change{u} \amp = \frac{1}{u'} \cdot (u - u') \\ \amp = \frac{1}{1 / s'} \cdot \frac{1}{s s'} \, \change{s} \\ \amp = \frac{1}{s} \change{s} \text{,} \end{align*}
which is exactly equal to the area of the rectangle in Figure 8.2.8.
🔗
By matching up areas of rectangles in this way, we see that the two upper sum approximations would compute to the same total approximate area. That is, every upper sum approximation for the integral defining \(\ln a\) corresponds to an upper sum approximation for integral (✶✶) with the same value. This would also work the other way — every upper sum approximation for integral (✶✶) could be “flipped” over by taking reciprocals of the rectangle base boundary points to create an upper sum approximation for the integral defining \(\ln a\text{.}\) And by changing sample points, we would see the same correspondence for lower sum approximations of these two integrals.
🔗
Therefore, the patterns of upper and lower sum approximations for each integral would “squeeze” down to the same values. That is,
\begin{equation*} \ccmint{1}{a}{\frac{1}{s}}{s} = \ccmint{1 / a}{1}{\frac{1}{u}}{u} \text{.} \end{equation*}
The first integral is the value of \(\ln a\text{,}\) while the second integral is the value of \(- \ln (1 / a)\text{,}\) using Property 2 of Convention 6.3.12. Flipping the negative sign to the other side of the equality of integrals above, we obtain
\begin{equation*} \ln (1 / a) = - \ln a \text{,} \end{equation*}
as desired.
🔗
🔗
We could rewrite Pattern 8.2.7 as
\begin{equation*} \ln (a^{-1}) = - \ln a \text{.} \end{equation*}
Using Pattern 8.2.4, we also have
\begin{gather*} \ln (a^2) = \ln (a \cdot a) = \ln a + \ln a \\ \implies \ln (a^2) = 2 \ln a \text{.} \end{gather*}
In both instances we see an exponent-becomes-coefficient pattern. Though we are not yet prepared to justify it, in fact this pattern holds for all exponents.
🔗
Finally, we can combine Pattern 8.2.4 and Pattern 8.2.7.
🔗
🔗
🔗
PrevTopNext