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Unit 3 Mega-assignmentSections 3.1 and 3.2
1. Given that logp2q “ 0.301, logp3q “ 0.477 and logp5q “ 0.699 (where these are base-10
logs), calculate
(a) logp6q
(b) logp600q
(c) logp25q
(d) logp1.5q (There are at least two ways to do this with what is given. Can you find
them?)
Section 3.3
2. The Richter scale defines the magnitude of an earthquake to be logpI{Sq (base-10 log)
where I is the intensity of the earthquake as measured by the amplitude of the waves traced
by a seismograph that is 100 km from the epicenter of the earthquake and S is the intensity
of a “standard” earthquake, one that moves the seismograph needle only 1 micron (which
is 0.0001 cm). The 1989 Loma Prieta earthquake in San Francisco had a magnitude of 7.1
on the RIchter scale. The 1906 “great” San Francisco earthquake was 16 times as intense.
What was its magnitude?
Section 3.4
3. Using only the logarithms in problem 1 above (and a calculator that can add, subtract,
multiply and divide), calculate:
(a) log2 p5q
(b) log6 p4q
(c) log3 p18q (observe that 18 “ 2 ¨ 32 , and think about what log3 p32 q is).
4. Solve the equations (use only the data in problem 1 and that arithmetical calculator):
(a) 5x “ 36
(b) log3 pxq “ 0.631
(c) 92x “ 10
Section 3.5
ż4
2x dx by considering an appropriate area.
5. (a) Calculate
0
ż 12
(b) Based on part (a) and the scaling rule, calculate
0
2
x dx.
3
Section 3.6
6. Explain why, for any positive numbers x and c, the area under the graph of y “ t2 between
1 and x is 1{c3 times the area under the graph between c and cx.
7. Using the numbers given in problem 1, together with lnp10q “ 2.3026, calculate lnp2q,
lnp6q, and lnp25q.
Section 3.7
8. If lnpyq “ 3 ln x ` 2.3026, then what is the equation that relates y to x (of the form y “
something involving x)?
9. If log10 pyq “ 2x ` 1.301, then what is the equation that relates y to x?
Section 3.5.1: A digression on areas and scaling
In this section, we’re going to take a little break from thinking about logarithms, and
make a digression into some geometric thinking about areas and how they change when we
make scalings in the x or y direction. We’re going to apply these ideas in the next section to
define and understand something called the “natural logarithm”, so this section does in fact
connect with our mathematical agenda, even if it doesn’t seem to at first.
We’re going to start by introducing a new concept. It’s a concept that combines geometry
and algebra in a peculiar way that is extraordinarily useful (and we will be using it a lot in
Unit 5 in our study of probability). It does have a peculiar name, though. It’s called the
definite integral.
One usually encounters definite integrals for the first time in Calculus, where you learn
lots of tricks (ahem, techniques) for calculating them. We’re not going to concern ourselves
with that part, we just want to have the concept and work with it geometrically (so we’re
using the mathematics as a language here).
Okay here we go: Suppose you have a function of a single variable, so in general maybe
it’s called f pxq. You can think of functions like x2 5 or 3x 2, most anything you like.
Then we’re going to draw the graph of f pxq, and choose two values of x, say x a and
x b with the property that a
b, f paq is positive and f pbq is positive and f pxq is also
positive for all x between a and b. Here’s a picture of a “generic” function with such a choice
of a and b.
The quantity we want to define is the area between the graph of f and the x-axis between
x
a and x b, which is shaded in the figure. The symbol for this concept looks like this:
»
b
a
f pxq dx and you say “the integral from a to b of f pxq”. Don’t worry about the dx for
now. And instead of that jargon, we’re often just going to say “the area under the graph of
f between a and b”.
2
» 2 So for example the area under the graph of x
2
x2
5 between x
2 and x
2 is
5 dx:
which happens to be a little more than 25 (don’t worry about
how I know that) and the area
»
under the graph of 3x
2 between x 0 and x 4 is
4
3x
2 dx:
0
Now this area you should be able to calculate: It’s a trapezoid, and you might remember the
area of a trapezoid (or maybe not). . . it’s also a triangle on top of a rectangle. The base of
the rectangle is the interval r0, 4s on the x-axis and the height of the rectangle is the interval
r0, 2s on the y-axis, so the area of the rectangle is 4 2 8. Then the base of the triangle
is the interval r0, 4s in the x direction along the line where y 2, and the height is the
interval r2, 14s along the line where x 4, so the area of the triangle is 21 4 12 24. So
altogether the total area is 24 8 32. And so we’ve worked our first integral!
»4
3x
0
2 dx 32.
Okay, so now we’ve got the idea of the definite integral. By the way, you may have
noticed that sometimes I’m calling this the definite integral and other times just the integral.
In fact, there are indefinite integrals as well, but we’re not going to be using them, so from
now on we’ll just call it the integral.
There is one convention and one additional observation regarding the integral that we are
going to use. The convention tells us says what to do in case b a. Since this means that
in some sense we’re measuring the area “backwards”, we’ll define
»a
b
f pxq dx
»b
a
f pxq dx
The additional observation is that there is nothing special about the variable x in the integral
— in many ways it is just like the index i in summation — sometimes these variables are
called “dummy variables” because they actually play no role in the process. So we could just
as well have said
»
»
4
3t
0
2 dt 32
4
3β
or
0
2 dβ
32
etc. You get the idea.
We will also need two properties of the integral that are straightforward from a geometric
point of view, even though one of them is a little bit complicated from the algebraic perspective. Both properties involve scaling — either reducing or enlarging a graph by a fixed ratio
— either along the x or y direction.
It turns out that the y direction is the easy one to understand. If you want to double the
distances in the y-direction, so that the height of the graph of a function f pxq above the
x-axis gets multiplied by 2, then you need only multiply the function by 2. The area under
the graph of y 2f pxq is twice as tall as the area under the graph of y f pxq. And since
the width doesn’t change, it stands to reason that the area will double, too. So
»b
a
2f pxq dx 2
»b
a
f pxq dx
Now there’s nothing special about doubling. If we multiply the function by any constant, say
k, then the area will get multiplied by k as well, since the heights will get multiplied by k
and widths will stay the same. To express this with our integral notation, we have
»b
a
kf pxq dx k
»b
a
f pxq dx
for any (positive) constant k.
The trick is doing stretches in the x-direction. Let’s do it for a specific, simple function
first, and then generalize what we learn. Let’s think about the part of the graph of the
function f pxq 2x 4 for x between 1 and 5. It’s the straight line segment between the
points p1, 6q and p5, 14q. If we wanted to stretch this horizontally by a factor of 3, then we
should move x 1 to x 3 and we should move x 5 to x 15 but otherwise keep all
the y’s the same — so we would have the straight line from the point p3, 6q to the point
p15, 14q.
What is the equation of this new line? Since x 0 would stay put in the stretching, the
y-intercept of the new line will be 4, just like for the old one. But the slope will be different:
p14 6q{p15 3q 8{12 2{3. So the new line is y p2{3qx 4.
And if you think about it — if you want to find the point on the new line for any particular
value of x, you first have to undo the stretching — that is, divide the x by 3 — and plug
this value into the original equation. So maybe it would be better to write the equation of
the new line as y 2px{3q 4.
And that’s the trick for any function — if you stretch the graph of y
by a factor of L then you get the graph of y f px{Lq.
f pxq horizontally
What happens to areas when you do this? Since horizontal widths are being stretched (or
compressed) by a factor of L, but vertical heights aren’t changing, the area will get multiplied
by the same factor, L. We can express this in our integral notation is
» Lb
f
La
x
L
dx L
»b
a
f pxq dx
Be sure to take your time unwinding exactly what this says, and what happened to all
the “moving parts” — for instance, in the stretched version, which is the left side of the
equation, we had to be sure to move the left and right endpoints from a and b to La and Lb
to be sure we get the corresponding areas. Try it with the part of the graph of y 2x 4
between 1 and 5 to see that it actually works.
We’ll let you practice a little bit with integrals and scaling, and then it’s on to the natural
logarithm in the next lecture.
Section 3.6.1: The natural logarithm
In this section, we’re going to continue our exploration of logarithms a bit further, and
arrive at a place that might seem scary at first, but it is of such importance in all parts of
math and science, particularly in data science, that it is absolutely essential that we come
to grips with it. Where we’re going to get is the definition of something called the “natural
logarithm”, and as I say, at first it’s going to seem that there’s nothing natural about it.
We could just jump in and define it, but I think that’s probably the worst thing to do,
since then there’s no motivation for the strangeness that’s going to happen. Instead, we’re
going to sneak up on it (rather as we did with logarithms themselves) so that maybe it will
in fact seem natural when we get there.
We’re going to recall with the basic properties of the logarithm function — or maybe now
we should say logarithm functions, since we’re getting used to using different bases for our
logarithms and we know that we write logb pxq for the power that you have to raise b to in
order to obtain x.
So log5 p25q 2; log2 p64q
that last one in?).
6; log10p0.001q 3 and logbpb8q 8 (see how we snuck
And we’ve solved equations with logarithms to different bases:
If 3x
12 then x log3p12q or if 22x 50 then 2x log2p50q and so x 21 logp50q.
Finally, we’ve converted from one base to another, particularly base 10, so we know that
log3 p12q
log10 p12q
log10 p3q
1.079
0.477
2.262
We know that all these different logarithm functions share four fundamental properties:
For any base b (¡ 0) and for any positive numbers x and y and any number p:
logb p1q 0,
logb pbq 1,
logb pxy q logb pxq
logb py q and
logb pxp q p logb pxq
In fact, the middle two properties, logb pxy q logb pxq logb py q and logb pbq 1 are the most
important ones, because the other two are consequences of these (and continuity).
And this works in reverse: If you have a function F pxq defined for all positive numbers
with the property that F pxy q F pxq F py q, then F is a logarithm function. The base of
the logarithm is equal to the number b that is the solution of the equation F pbq 1.
We’re going to use that observation, together with our work on areas and integrals, to
define a function that will turn out to be a logarithm even though there are no exponents or
logarithms in sight to begin with. The function is going to be called Apxq, and its definition
is as follows:
»x
1
dt for x ¡ 0,
Apxq
1 t
so Apxq is the area under the graph of y
1{t for t between 1 and x.
In this picture you can see the graph of 1{t, and the area that is the value of Apxq for a some
value of x ¡ 1. A few properties of Apxq should be clear from the picture:
1. If x gets bigger, there will be more area, so Apxq is an increasing function.
2. Ap1q 0, since if x 1 there is no width to the region, so no area.
3. If x
1, then Apxq
0 because we’ll be integrating “backwards” to go from 1 to x.
4. Because 1{t gets smaller and smaller as t gets larger and larger, the rate at which Apxq
increases gets less and less as x gets larger and larger.
All of these are properties of logarithm functions as well. So if we can show that A satisfies
the all-important property Apxy q Apxq Apy q, then A will in fact be a logarithm. How
can we show that A has this property?
This is where our experience with scaling areas is going to pay off. Let’s choose a scale
factor, call it c, we recall that if we rescale the y-axis by multiplying the height of everything
by c, then the area of the scaled version of region will be c times the area of the original
region. We can state this as
»x
1
c
dt c
t
»x
1
1
dt cApxq.
t
We also know that rescaling the x-axis by multiplying all horizontal distances from x 0 by
c, then areas get multiplied by c as well:
» cx
c
1
dt
t{c
» cx
c
dt c
t
c
»x
1
1
dt cApxq.
t
Here’s the inspired part: what we’re going to do is rescale the x-axis by multiplying by
c and simultaneously rescale the y axis by dividing by c (or, equivalently, by multiplying
by 1{c). This will leave areas unchanged since multiplying by 1{c undoes multiplying by c.
But something remarkable will happen. The effect of the two rescalings is captured in the
following integral:
»
1 cx c
dt
c c t
But we can manipulate this integral by using our y-rescaling rule:
1
c
» cx
c
» cx
c
dt
t
c
1c
dt
c t
» cx
c
1
dt
t
And this is an amazing property of the function 1{t (in fact, it is a property that only constant
multiples of 1{t among all functions possess). So we’ll memorialize this as follows:
Amazing property of 1{t: For any positive numbers x and c, the area under the graph of
1{t between 1 and x is the same as the area under the graph between c and cx.
And this is just what we need to get the log property for A: Given our two numbers c
and x we know that
»c
»x
1
1
Apcq
dt and Apxq
dt
1 t
1 t
But the amazing property of 1{t tells us that
Apxq
»x
1
1
dt
t
» cx
c
1
dt.
t
This implies that
Apcq
Apxq
»c
1
1
dt
t
» cx
c
1
dt
t
» cx
1
1
dt Apcxq
t
as you can see from the graph (the area from 1 to c ends where the area from c to cx starts,
so their sum is just the area from 1 to cx):
Great! So this function Apxq is actually a logarithm. I suppose it’s reasonable to call this
logarithm the “natural” one since it comes up naturally in a context that has nothing to do
with logarithms (after all, we thought about the graph of 1{t way back in Unit 1).
But what is its base? Well, we know that the base of this logarithm is the number b
for which Apbq 1. That’s the point on the horizontal axis for which the area under 1{t
between 1 and b is 1. We can’t find it precisely, but we can estimate it as follows:
First, it’s pretty clear that this number b is bigger than 2, because the area under 1{t for
t between 1 and 2 fits inside the 1-by-1 square with base r1, 2s on the t axis and height
r0, 1s on the y-axis and leaves a lot of area left over. So Ap2q 1.
We can also show that b is less than 4, because we can take three rectangles: the first has
base r1, 2s and height r0, 1{2s, the second has base r2, 3s and height r0, 1{3s and the third has
base r3, 4s and height r0, 1{4s — these three rectangles don’t overlap, and they all fit inside
the area that defines Ap4q — the total of the areas of the rectangles is 21 13 14 13
¡ 1,
12
so Ap4q ¡ 1.
Using variations of this idea with more and more little rectangles, you could show that
Ap2.5q
1 and Ap3q ¡ 1 (you need at least 8 little rectangles to do these). And using
what Archimedes called the “method of exhaustion” you can finally show that the number
for which A 1 is approximately 2.71828 (it’s actually an irrational number). This number
is called e in honor of Leonhard Euler and it is the base of the natural logarithms.
Rather than loge pxq for the natural logarithm of x, it is standard to write lnpxq. And
even though it’s called “natural logarithm”, and you might think that “ln” is the abbreviation
for “natural log” in some language like French that puts the adjectives after the nouns, it’s
actually the abbreviation for “logarithme napierienne”, in honor of John Napier who invented
the logarithm at the beginning of the 17th century.
Now you’ve got to be thinking, “Why would anyone use such a cockamamie base for their
logarithms? And how are we ever going to be able to calculate them?”
The answer to the first question comes from the definition of lnpxq via areas — our
function Apxq. The way we were estimating areas by adding up the areas of little rectangles
makes it possible to calculate values of lnpxq very precisely.
And once we have values of lnpxq, it’s easy to convert them to values of logpxq log10 pxq
— remember, it’s just multiplying or dividing by a conversion factor, and the conversion factor
is lnp10q 2.3026 . . . If you know log10 pxq then you multiply it by 2.3026 to get lnpxq, and
if you have lnpxq you divide it by 2.3026 to get log10 pxq.
Finally, you could ask why we don’t make an area definition for logpxq? Well, we can, but
it would use areas under the graph of 1{p2.3026 tq from 1 to x. As things turn out, there’s
got to be a weird number somewhere, and over the centuries mathematicians and scientists
have agreed to make it the base of the natural logarithm. And in fact, you’ll get used to
using lnpxq and the exponential function ex as we continue our work in this course.
