Logarithms are one of the
more difficult concepts
in the precalculus curriculum.
They're a difficult idea
for several reasons.
The first reason is the
definition is difficult.
And even then, once
you kind of understand
the meaning of logarithms, the
process of calculating them
is very challenging as well.
So what I'm going
to do to begin with
is to define
logarithms carefully
and talk about how you can find
them in certain circumstances.
So let's remind ourselves
how we started off
thinking about exponents.
So the first thing
we thought about
was, if I took my
basic operations, so
addition and multiplication,
and I repeatedly did them.
So let's think about repeated
addition for a second.
So repeated addition-- repeated
addition is just the same
as multiplication.
So just as an example--
as an example, if I was
to take 3 and add it
to itself four times,
so four copies of this,
we know this is
equal to 4 times 3,
basically by definition of
multiplication pretty much.
This is equal to 12.
OK.
So what's the reverse
process of this?
Well, the reverse
process is what?
It's the reverse
of multiplication.
And the reverse of
multiplication is division.
So let's just kind
of think about this.
So the reverse process
must be the inverse process
to multiplication,
which is division.
Now, what do you mean
by reverse process?
Well, I think up
here, let's kind of--
let me explain what I mean.
How do I go from 4 to 12, right?
That's kind of the
forward process.
Well, I would multiply 4
by 3, and I would get 12.
So the reverse process would
be doing this backwards,
essentially.
So let's take a moment to
think exactly what that means.
So imagine I'm adding 3 to
itself some number of times.
So I don't know
exactly how many,
but let's label this
as a times, where
a is going to be some positive
whole number, let's say.
Now, let's imagine this
is equal to 15, OK?
So the reverse of this is
basically asking, well,
how do I go backwards?
How do I go from 15 to a?
What's the procedure?
So as I just described,
it's got to be division.
Let's carefully check that.
So this implies that 3
times a has to be equal to--
or a times 3, it
doesn't matter how
you write it-- is equal to 15.
And then one of my
basic laws of algebra
says this implies that a is
equal to 15 divided by 3.
But that's, of
course, equal to 5.
And just to really
be careful, what
am I actually invoking here?
I'm invoking the property that
ab equals c implies that a is
equal to, let's say, b divided
by-- sorry, c divided by--
c divided by b.
So this is the kind of a
general property of algebra
I'm using, OK?
So you could have an if
and only if condition
assuming b is non-zero.
You got to be a
little bit careful,
otherwise things
can go a bit wrong.
And here, b is 3,
so it's non-zero.
Very good.
So I'm just being really
careful about making
sure every step is a very
simple principle of algebra.
All right.
So that's addition, right?
Repeated addition
is multiplication.
The reverse process is the
reverse of multiplication,
which is division, right?
A familiar operation.
What about if I
repeat multiplication?
So repeated multiplication is
what I called exponentiation.
So just as an example--
as an example, let's
say 2 times itself--
[INAUDIBLE] now let's go to 6.
So I multiply 2 by
itself six times.
2 multiplied by
itself six times,
well, that's equal to
2 to the 6, of course.
It's the notation
I'm used to using.
What's that actually equal to?
2 to the 6 is 64.
So let's kind of do an analogy
with what's happening above.
In a sense, what I'm doing
is I'm going from 6 to 64
by raising 2 to
the power 6, right?
So the forward process,
if you want, is this.
OK?
So I go from 6 to 64 by
raising 2 to that power.
Very nice.
Now, what's the reverse process?
The reverse process
would be going backwards.
It would be essentially
going from 64 back to 6.
So this reverse process is
the reverse of exponentiation,
and that's not something
we've encountered before.
Previously, it was
multiplication,
so the reverse was
definitely division.
We are very used to that.
But the inverse
of exponentiation
is something new.
So that's what logarithms are.
They're the inverse
process of exponentiation.
So I will go through a
few examples in a second,
of course.
Logarithms.
So logarithms is
exponentiation in reverse.
So this is, in a
sense, a new operation.
And we're going to see
it's a lot more complicated
than division.
It's more difficult.
But it's something
that we're going to have
to kind of get used to,
because it's going to be one
of our core new operations
on functions, if you want.
So let's do two
examples, I think, just
to make sure we understand what
this reverse process actually
means.
So let's do exactly what we did,
really, with the sum example,
but we'll do it
with multiplication.
So 2 times 2, dot, dot,
dot, some number of times.
Let's say a times.
Let's imagine this is
equal to 32 for a second.
So what does this mean?
Well, this would
imply that, well,
2 multiplied by itself a times
in my usual exponent notation
is 2 to the a has
to be equal to 32.
OK?
So what we're really
trying to do here
is the reverse of this
exponentiation process.
So we're trying, in a
sense, to go backwards.
So we're trying
to work out what's
in the opposite direction.
So I'm looking for
some a that does this.
Now thankfully, there is
a pretty familiar a to us
that does it.
So observation-- I mean,
2 to the 5 is equal to 32.
That's true.
So it turns out that's the only
number that will work, right?
That is the only exponent
which will give me 32
if the base is 2.
So there's one option--
a has to be equal to 5.
OK.
So that was OK, but
it wasn't great,
because I needed to spot that
2 to the 5 is equal to 32.
Let's do another example.
So let's do 3 times
itself a bunch of times.
Let's do the same deal.
Let's label this as a.
And let's say this is
equal to, I don't know, 81.
So it's equal to 81.
So what we're trying to
do is the reverse process
of exponentiation.
We're trying to work out
what a would do the job.
Same story.
So if this happened,
we would have 3
to the power a is equal to 81.
Does that look like
anything we've seen before?
Well, let's just go
through some small numbers
just to see what
happens, shall we?
So if I do 3 to the power
1, of course, that's 3.
If I do 3 squared,
that's equal to 9.
If I do 3 cubed,
that's equal to 27.
If I do 3 to the 4,
that's equal to-- bingo--
81.
So all of this tells me,
essentially by trial and error
more than anything else, I have
a is equal to 4 as a solution.
And it turns out that's
the only possible,
as we'll see in a moment.
Right.
So what I want is some
notation here, right?
I want to have some notation
which talks about a being,
in a sense, the opposite
of exponentiation.
So that's what
logarithm notation is.
So maybe I'll leave this
up and just do it below.
So, notation.
So to represent the
relationship between the output
here, 81 or 32, and the
number of terms, the exponent,
we use log notation.
So the first guy I
would write log base 2--
because 2 is the base--
of 32 is equal to a.
So a is equal to 5 here.
So as with
exponential functions,
or exponential exponents,
the thing in the bottom
is called the base.
Same thing for the logarithms.
Now what about the second guy?
Well, it would be the statement
that log to the base 3 of 81
is equal to 4.
So what are these shorthand for?
They're basically shorthand for
the following two statements.
This is just alternate
notation for the statement
that 2 to the 5 is equal to 32.
That's all it's saying.
It's just a funny
notation for it.
This is the same deal.
This is just saying that
3 to the 4 is equal to 81.
It's just new notation
when it comes down to it.
So what should this
notation mean in general?
So essentially,
what we want to do
is have a relationship between--
have a relationship between
the base and the output, right,
the exponentiation.
So let's think about what
the general case means.
So the general case.
So what have we got?
Well, I've got some base to
some power a is equal to c.
And I imagine I know
what b is and I know what
c is, right, just like here.
I know 2 and 32 and
3 and 81, and I need
to know what the exponent is.
That's the a.
So the notation I'm going
to use for this is--
so if and only if, the
notation I'm going to use
is log to the base of
b of c is equal to a.
So that is going to be
the meaning, if you want,
of what log to the
base b of c is.
So this is very important.
This is basically me defining--
this is me defining
what a logarithm means.
Ah, if it plays nice.
This is me defining
what log notation means.
So when you say log to
the base b of c is a,
it literally is the
same thing as this.
This is the meaning
of the notation.
So I always find
this very confusing.
Certainly, when
I first saw this,
it was like, oh, my goodness,
this is so confusing,
it's all backwards, I'm
not used to exponentiation
done backwards.
But this is a definition, so
let me kind of write this here.
So again, as I've
said previously,
the definition means this is the
meaning of the notation, right?
There's no where-- we can't
go any further back than this.
This is it, right?
Now, I do want to say a
few things about this.
First of all, we don't
want b to be negative,
because if b is
negative, we have
terrible problems with powers.
So let's just completely
ignore this issue.
We don't want to define
any logarithms or anything
like that when b is negative.
So we'll assume b
is greater than 0.
Next, we don't want
b to be equal to 1.
The reason for that
is, if b is equal to 1,
I mean, 1 to any
power is just 1,
so this would be
very meaningless when
it comes down to it, right?
So we don't want to include 1.
That doesn't make sense, OK?
Basically, 1 to any power
is equal to 1 for all a.
OK.
So the meaning of log, you
need to, at the very least,
b has to be strictly positive,
and it can't be equal to 1.
Next, a little bit
more notation--
a little bit more notation here.
Often people drop the base.
And if you drop the
base and just write log,
all that means is
log to the base 10.
Now, the reason that's natural
is because log to the base 10,
it deals with base 10, and 10
is one of the most commonly
used bases.
Remember, decimals
are all in base 10,
so it makes sense that
this is somehow privileged.
You'll find out when
you go on to do calculus
there are other bases which are,
in many ways, more important
than base 10.
But for the moment, it's worth
thinking about them in general.
A few special
cases as well, just
to make sure that we
really understand this.
And we'll talk about intuition
for counting these in a second.
But really simple cases
are the following.
Well, b to the 1 is
equal to b, right?
Of course.
So that implies-- what it
implies is that log to the base
b of b is equal to 1
by definition, right?
Similarly, b to the
0 is equal to 1.
So that would imply log to
the base b of 1 is equal to 0.
So these are just coming from
the definition, if you want.
So really what I want
to say about this
is a little bit about
calculation, first of all.
So there's an intuition
that goes into this.
This is a very
confusing statement.
It's like, oh, my goodness,
what am I-- what am I
meant to do here
to calculate this?
Here's how I think about it.
The intuition is, I mean,
what do we need to raise b to
in order to get c?
That's kind of the
question I ask myself--
to raise b to in order to get c.
That's kind of the thing
hardwired into my head.
I say, OK, what power of b
do I need in order to get c?
Now, we know powers
can be anything now,
if I'm assuming b is positive.
Remember, b to the a makes
sense for any number at all.
So there's no reason the
output of the logarithm
has to be a positive number.
So let's do a
couple of examples--
couple of examples.
First, let's do log
to the base 9 of 3.
So what's this equal to?
So what do we need?
So we need an a, such that 9 to
the power a-- so let me write
that more carefully--
such that 9 to the
power a is equal to 3.
Well, what works?
Thankfully, I know the square
root of 9 is equal to 3,
so 9 to the power
1/2 is equal to 3.
So this will imply--
well, this actually guarantees
that log to the base 9 of 3
would have to be equal to what?
It would have to
be equal to 1/2.
That's actually the only
number that's going to work,
as we're going to
see in a second.
So the logarithm
doesn't have to be--
doesn't have to
be a whole number.
So that's the
first thorny issue.
Let's do another one beside it
to illustrate another issue.
What about, let's say,
log to the base 4 of 1/4?
What does that mean?
Well, again, we need to
find an a such that--
OK, what's it going to be?
It's going to be 4 to the
power a has to be equal to 1/4.
Thankfully, this is a very
obvious example, right?
This is my standard reciprocal.
4 to the minus 1
is equal to 1/4.
So what does that tell you?
It's going to tell you that
log to the base 4 of 1/4,
it's going be minus 1.
So interesting, the output of
a logarithm could be negative.
And that's pretty
obvious, really,
because this thing here
can be any positive.
Well, it can be
any number at all,
positive or negative, right,
as long as b is strictly
greater than 0.
OK, so interesting.
So these examples were
only really doable by hand,
because we could spot
a power that worked.
The problem is, we can't
do that in general,
and this really is
what makes logarithms
so difficult on a first sweep.
So problem, at least
for calculating things,
it could be very hard--
could be very hard to
find exact value by hand.
So I'll give you a little
bit of intuition here.
Imagine if I asked
us, I don't know,
let's do log to the base
2 of 40, and I asked us,
what's that equal to?
So I'm going to
sit down and just
try to work out powers of 2
and see what could be going on.
So 2 to the 1 is
equal to 2, of course.
2 to the 2 is equal to 4.
2 to the 3 is equal to 8.
2 to the 4 is equal to 16--
getting bigger.
2 to the 5 is equal to 32.
OK, not far off 40.
But then unfortunately, 2 to the
6 jumps all the way up to 64.
So intuitively-- this is
not precise mathematics--
it seems to me that
the a I'm looking
for is going to be
somewhere between 5 and 6.
Because as I move from
5 to 6, the exponent
is going to move from 32 to 64.
At some point, it's
going to hit 40.
So that's not precise, but it
is correct logic it turns out.
So there is some number--
some number between,
let's say, 5 and 6.
So some number-- I'll
just say some a--
such that 2 to the
a is equal to 40.
And that's what
we're looking for.
That would be the logarithm.
But it's difficult
to get your hands on.
So you can see, first
of all, logarithms
not easy to understand,
difficult to calculate,
and also there are certain
circumstances where they
don't really make any sense.
So let me talk a little bit in
general about what logarithms
look like and when they exist.
So here's some important facts.
OK, important facts.
So again, we're going to assume
b is strictly greater than 0
and is not equal to 1.
Otherwise, it
doesn't really work.
OK, so let's think
about log sub b of c.
Now, does this make sense?
For it to make sense, you'd
need to be able to write down
c in the form 2 to the power a.
Now, it turns out that
can be always done
if c is greater than 0, right?
Remember, c is equal to
b to the a in theory.
Turns out you can always do it.
So we can always find--
and actually a unique a.
There's only one
number that does it.
I've already kind of
alluded to this little bit--
a unique a such that b
to the a is equal to c.
So this tells you, great,
the logarithm makes sense,
there's only one choice
for the logarithm.
All good, right?
So the log bc is defined.
Of course, it's equal to a.
There's only one choice.
Fantastic.
For example, I don't know,
2 to the power of something
is equal to 4.
Well, the something in
the exponent has to be 2.
There's no other option.
But if it was 2 to the power of
something was equal to 1,000,
there'll also be some number.
It won't be easy to
get your hands on it,
but it'll be there somewhere.
And that would work
with smaller numbers,
too, as long as
they're positive.
For example, 2 to
the power of 1/2--
well, sorry, 2 to the power
of something equals 1/2,
that something would be minus
1, like we saw a second ago.
2 to the power of
something is equal to 1/4,
that would be minus 2.
So this a could be negative,
and the more negative
it is the more--
the smaller the value
of c, if you want.
OK, what about if c is
negative or indeed equal to 0?
Well, then it's no good, right?
Remember, one of my
properties of exponentials
was the fact that if b is
positive, b to any power
is always strictly positive.
So b to the a is always greater
than 0, strictly greater than 0
for all numbers a.
So this implies b to the
a can never equal c, ever.
So that means no a will work,
so that tells me log sub b of c
is undefined.
It's undefined,
doesn't make sense.
So the takeaway is if c is
strictly greater than 0,
there's a logarithm,
there's only one.
If c is less than or equal
to 0, it doesn't make sense.
So just as a quick
example here--
so example, if we had a
scenario with c is minus 1.
So 3 to the x, let's
say the base is 3,
this has no solutions, right?
3 to any power is
always positive.
I'm only using x here,
because we're just
used to solving for x.
But it has no solutions,
so this tells you
that log to the base 3
of minus 1 is undefined,
doesn't make sense.
So it's kind of quite
subtle general properties.
So the last thing
I want to do is
to talk about an important
simplifying principle
when you're dealing with--
when you're dealing with
logarithms or exponents.
Actually, maybe I should write
a conclusion down for this.
So let's just write
this as a conclusion.
I think that might
be a good idea.
So the conclusion of what
I just said basically
is the following--
log sub b of c is defined if
and only if c is greater than 0.
And again, when I'm writing
this down, I'm always assuming--
I'm always assuming
that b is not 1
and it's strictly
greater than 0.
So b has to be bigger than
0, and b is not equal to 1.
It can't be 1, because 1
to any power is just 1.
So that's kind of the
conclusion of what
I've just said in this
small, complicated diagram.
So finally, a very
important property
which will allow us to
manipulate equations
with exponents and logarithms
is the cancellation property.
So we'll get a lot more practice
with this in further videos.
So calculation properties--
cancellation properties.
So these are true basically
by definition of log,
but it's worth
writing them down.
If I take b and I raise it
to the power log b of c,
this is just equal
to c again, OK?
Of course, it's the
definition of log.
[INAUDIBLE] move this
down for more space.
And we could also take the log
of-- kind of an exponential log
sub b of b to the a.
It's just equal to a
basically by definition.
So these are the
cancellation properties.
They allow us to, in a
sense, cancel logarithms
with the first, if I
raise them to a power,
and cancel an exponential,
or an exponent,
if I take the log of it.
So we'll get more practice
with these in further videos,
as I said.
Worth saying here, of course,
c has to be greater than 0,
b has to be greater than
0 and not equal to 1,
and a can be anything.
Remember, you can take-- b to
any power of b is positive.
So these are the
cancellation properties.
Let's do a quick example,
a numerical example.
I'm not going to go into
complicated examples yet.
But let's just make sure
it's true, shall we?
So 3 to the power
3 is equal to 27.
So this tells you what?
It tells you that log sub
3 of 27 is equal to 3.
OK, so let's just
check both sides
and show they're
equal to each other.
So this implies that 3
to the power log 3 of 27
is equal to 3 to the power
3, which is equal to 27.
Tick, that is the first
property, exactly the same.
And of course, the second one
is kind of totally obvious.
Log-- well, it's
not obvious obvious,
but it's just true by
definition-- log with a base
3 of 3 to the 3 is just 3.
Or maybe I should just say,
to really hammer it home,
this is log to the base 3 of 27.
And as before, you've
seen that it's equal to 3.
So these are both true.
So to recap, logarithms
are essentially the inverse
of an exponentiation.
So this is the crucial defining
property of logarithms.
They're difficult to calculate.
In general, you can't
really do it by hand.
And their existence is
a little bit delicate.
It all depends on
your b and your c.
We always assume
b is truly greater
than 0 and b is not equal to 1.
That's part of the
assumptions the whole time.
But you can't take log
of a negative number,
because b to any power
of b is positive.
It's always positive.
Whereas a logarithm can
take any value as an output,
if you want.
a can be anything at all.
So this value can be
anything, because we
can raise b to any
power we want--
negatives, positives, whatever.
That's what logarithms are.
Next time, we'll look
at laws of logarithms.
