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Chapter 11 The natural logarithm

Section 11.1 Definition

Pattern 10.4.10 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 11.1.1. 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 11.1.2. The natural logarithm as accumulation under \(r(t) = 1/t \text{.}\)
In Figure 11.1.2, 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 11.1.2(b) shows negative accumulation values for \(0 \lt t \lt 1 \text{.}\) This is because of Property 2 in Convention 9.3.12: 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 11.1.3. The natural logarithm as accumulation under \(r(t) = 1/t \text{.}\)

Example 11.1.4. 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 10.4.14, since the integrand function is undefined at \(t = 0 \) we will “re-base” our integral to \(t = 1 \) (using a version of Pattern 9.3.14), and apply Formula 2 from Pattern 10.4.10 for the linear part of the integrand function.
\begin{align*} \ccmint{3}{4}{ t + \frac{1}{t} }{t} \amp = \ccmint{1}{4}{ t + \frac{1}{t} }{t} - \ccmint{1}{3}{ t + \frac{1}{t} }{t}\\ \amp = \left( \ccmint{1}{4}{t}{t} + \ccmint{1}{4}{ \frac{1}{t} }{t} \right) - \left( \ccmint{1}{3}{t}{t} + \ccmint{1}{3}{ \frac{1}{t} }{t} \right)\\ \amp = \left( \frac{4^2 - 1}{2} + \ln(4) \right) + ( \frac{3^2 - 1}{2} + \ln(3) ) \end{align*}

Remark 11.1.5.

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 11.1.1 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 11.2 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 11.1.2(b) or Figure 11.1.3(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 9.3.11, 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{equation} \ccmint{a}{a b}{\frac{1}{u}}{u} \text{?}\tag{✶} \end{equation}
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 9.3.3). 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 11.2.5. 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 11.2.6.

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 11.2.8. 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{equation} \ccmint{1 / a}{1}{\frac{1}{u}}{u} \text{,}\tag{✶✶} \end{equation}
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 11.2.9. 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 11.2.9 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 11.2.8.
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 9.3.12. 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 11.2.7 as
\begin{equation*} \ln (a^{-1}) = - \ln(a) \text{.} \end{equation*}
Using Pattern 11.2.4, 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 11.2.4 and Pattern 11.2.7.

Example 11.2.12.

Part of Example 11.1.4 involved computing the definite integral
\begin{equation*} \ccmint{3}{4}{\frac{1}{t}}{t} = \ccmint{1}{4}{\frac{1}{t}}{t} - \ccmint{1}{3}{\frac{1}{t}}{t} = \ln(4) - \ln(3)\text{.} \end{equation*}
We can now use Pattern 11.2.11 to “simplify” this result to a single logarithm:
\begin{equation*} \ccmint{3}{4}{ \frac{1}{t} }{t} = \ln(4 / 3) \text{.} \end{equation*}
In light of Example 11.2.12, we record the following version of Pattern 11.2.11.

Section 11.3 Derivative of the natural logarithm

Subsection 11.3.1 Initial formula

In Section 10.5, we found that
\begin{equation*} \opddt \, \ccmint{0}{t}{u^m}{u} = t^m \end{equation*}
is true for all positive exponents \(m \text{.}\) In Checkpoint 10.5.1, you should have also found that
\begin{equation} \opddt \, \ccmint{1}{t}{u^m}{u} = t^m \tag{†} \end{equation}
is true for all negative exponents \(m \neq -1 \text{.}\) We will explore this integration-then-differentiation pattern more generally in Chapter 14, but for now we will take it on faith that pattern (†) holds even in the case of exponent \(m = -1 \text{:}\)
\begin{equation*} \opddt \, \ccmint{1}{t}{u^{-1}}{u} = t^{-1} \text{.} \end{equation*}
Since the integral function being differentiated here is precisely the natural logarithm, we have a new differentiation pattern.

Example 11.3.2. Derivative involving a logarithm.

For
\begin{equation*} q(t) = t^2 + \ln(t) + 6 \text{,} \end{equation*}
we have
\begin{align*} q'(t) \amp = \opddt \bbrac{t^2 + \ln(t) + 6} \\ \amp = \ddt{t^2} + \ddt{\bbrac{\ln(t)}} + \ddt{6} \\ \amp = 2 t + \frac{1}{t} + 0 \end{align*}
where in the first step we have applied the Sum Rule and in the second step we have applied the Power Rule, our new derivative formula for the natural logarithm, and the rule for the derivative of a constant.

Example 11.3.3. Derivative of a scaled logarithm.

Let \(a,b \) represent constants and consider \(q(t) = a \ln(b t) \text{.}\) As this function is both vertically and horizontally scaled, we compute its derivative by applying the Constant Multiple Rule and also Pattern 6.3.17.
\begin{align*} q'(t) \amp = \opddt \bbrac{ a \ln(b t) } \\ \amp = a \ddt{\bbrac{ \ln(b t) }} \\ \amp = a b \ln'(b t) \\ \amp = \frac{a \bcancel{b}}{\bcancel{b} t} \\ \amp = \frac{a}{t} \end{align*}
Alternatively, we could apply properties of the logarithm before differentiating:
\begin{equation*} q(t) = a \ln(b t) = a \bbrac{ \ln(b) + \ln(t) } = a \ln(b) + a \ln(t) \text{.} \end{equation*}
By the Sum Rule, we can now differentiate the two terms separately and add the results. As \(a \) and \(b \) are constants, the expression \(a \ln(b) \) is also constant, so the derivative of the first term is \(0 \) by our rule for the derivative of a constant. Then in the second term we need only apply the Constant Multiple Rule.
\begin{align*} q'(t) \amp = \opddt \bbrac{ a \ln(b) + a \ln(t) } \\ \amp = \opddt \bbrac{ a \ln(b) } + \opddt \bbrac{ a \ln(t) } \\ \amp = 0 + a \ddt{\bbrac{ \ln(t) }} \\ \amp = a \cdot \frac{1}{t} \\ \amp = \frac{a}{t} \end{align*}

Subsection 11.3.2 Extended formula

It is common to extend the domain of the natural logarithm by composing with the absolute value function: whereas \(\ln(t) \) is defined only for \(t \gt 0 \text{,}\) \(\ln\abs{t} \) is defined for all \(t \) except at the singularity at \(t = 0 \text{.}\) For this function, a negative input produces the same output value as the corresponding positive input of the same magnitude, so the graph is symmetric about the vertical axis.
Graph of the natural logarithm composed with the absolute value.
Figure 11.3.4. Graph of the quantity function \(q(t) = \ln\abs{t} \text{.}\)
Consider a point on the graph in Figure 11.3.4 at \(t = t^\ast \) with \(t^\ast \lt 0 \text{.}\) There is a corresponding point on the other side of the vertical axis, at input value \(t = \abs{t^\ast} \) but at the same output height. Since \(\abs{t^\ast} \gt 0 \text{,}\) this second point is on the graph of \(y = \ln(t) \text{,}\) and so we know its derivative value is the reciprocal of the input value for that point:
\begin{equation*} \funceval{\ddt{\bbrac{\ln(t)}}}{t = \abs{t^\ast}} = \frac{1}{\abs{t^\ast}} \text{.} \end{equation*}
This derivative value describes the slope of the tangent line at this point.
On the other hand, the derivative value
\begin{equation*} \funceval{\ddt{(\ln\abs{t})}}{t = t^\ast} \end{equation*}
describes the slope of the tangent line at the original input value \(t = t^\ast \text{.}\) But since these two points are reflections of each other in the vertical axis, we can see that the two slopes will be equal in magnitude but have opposite signs.
Two tangent lines at symmetric points on the graph of the natural logarithm composed with the absolute value.
Figure 11.3.5. Two tangent lines at symmetric points on the graph of the quantity function \(q(t) = \ln\abs{t} \text{.}\)
So by symmetry we have
\begin{equation*} \funceval{\ddt{(\ln\abs{t})}}{t = t^\ast} = - \frac{1}{\abs{t^\ast}} = \frac{1}{- \abs{t^\ast}}\text{.} \end{equation*}
However, since \(t^\ast \) was negative to begin with,
\begin{equation*} - \abs{t^\ast} = t^\ast \text{,} \end{equation*}
and we end up with
\begin{equation*} \funceval{\ddt{(\ln\abs{t})}}{t = t^\ast} = - \frac{1}{\abs{t^\ast}} = \frac{1}{t^\ast}\text{,} \end{equation*}
which is the same reciprocal-of-input-value formula as for the derivative of the natural logarithm.

Example 11.3.7. Derivative of a horizontally transformed logarithm.

To calculate
\begin{equation*} \opddt \ln\abs{3 t - 5} \text{,} \end{equation*}
start with \(p_1(t) = \ln\abs{t} \text{.}\) Then Pattern 11.3.6 says that
\begin{equation*} p_1'(t) = \frac{1}{t} \text{.} \end{equation*}
Now consider
\begin{equation*} p_2(t) = p_1(t - 5) = \ln\abs{t - 5} \text{.} \end{equation*}
Pattern 6.3.15 says that
\begin{equation*} p_2'(t) = p_1'(t - 5) = \frac{1}{t - 5} \text{.} \end{equation*}
Finally, consider
\begin{equation*} q(t) = p_2(3 t) = \ln\abs{3 t - 5} \text{,} \end{equation*}
the function whose derivative we would like to calculate. Pattern 6.3.17 tells us
\begin{equation*} q'(t) = 3 p_2'(3 t) = \frac{3}{3 t - 5} \text{.} \end{equation*}
So we have determined that
\begin{equation*} \opddt \ln\abs{3 t - 5} = q'(t) = \frac{3}{3 t - 5} \text{.} \end{equation*}
Recall that the natural logarithm is defined to be the accumulation function \(A_1(t) \) for rate function \(r(t) = 1/t \text{.}\) That is, values of \(\ln(t) \) are defined to be results of definite integral calculations. Using symmetry, we can interpret values of \(\ln\abs{t} \) the same way.
Two corresponding areas under the graph of the reciprocal function.
Figure 11.3.8. Two corresponding areas under the graph of \(r(u) = 1/u \text{.}\)
By symmetry, the areas in Figure 11.3.8 have the same magnitude, but are oppositely oriented:
\begin{equation*} \ccmint{t}{-1}{\frac{1}{u}}{u} = - \ccmint{1}{\abs{t}}{\frac{1}{u}}{u}\text{.} \end{equation*}
Using Property 2 of Convention 9.3.12, we can counteract the negative orientation by integrating backwards in time:
\begin{equation*} \ccmint{-1}{t}{\frac{1}{u}}{u} = - \ccmint{t}{-1}{\frac{1}{u}}{u} = \ccmint{1}{\abs{t}}{\frac{1}{u}}{u}\text{.} \end{equation*}
This leads to the following piecewise interpretation of the function \(\ln\abs{t} \text{.}\)
\begin{equation*} \ln\abs{t} = \begin{cases} \displaystyle \ccmint{1}{t}{\frac{1}{u}}{u} \amp t \gt 0 \\[3pt] \displaystyle \ccmint{-1}{t}{\frac{1}{u}}{u} \amp t \lt 0 \end{cases} \end{equation*}
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