>> Hi there.
This is Professor
Shannon Gracie.
We are working out of the
"Precalculus:
Enhanced with Graphing
Utilities" text by Sullivan
and Sullivan,
and today we'll be covering
properties of logarithms.
Let's warm it on up.
We've got a couple
of problems here
and the first one is,
show that log base A
at 1 equals 0, so go ahead
and pause the movie
and warm it up.
All right,
let's see how you did.
If we take a look
at this problem,
log base A at 1 equal to 0,
let's switch it
into exponential form.
This is true if and only
if so the base grows at tall,
a logarithm is an exponent,
so what's our power?
Awesome. Zero.
And what's left is though
equals 1, so what do we get?
A to the 0 is 1.
1 does indeed equal 1,
so you showed
that that is true.
OK, let's see
about the next one.
Again, going
from logarithmic form
to exponential form,
so the base grows up tall,
a logarithm is an exponent
and then what's left,
so A does indeed equal to A
and we have shown these
two properties.
So in summary,
what have found?
We found that log base A
at 1 is equal to 0
and log base A
at A is equal to 1.
So let's go
on to some general properties
of logarithms.
Now, I apologize ahead
of time, I must have been
tired when I was making the
notes and so there might be
some changes.
If you're working
with a revised edition,
I've probably made the changes
already but bear with me,
we'll make some changes
as we go along.
So let's let M
and A be positive real numbers
with A not equal to 1,
and let's let R be any
real number.
So first property here,
if you evaluate an exponential
expression base A
at a logarithm
of the same base,
you get what's inside the log
because you're evaluating a
function at its inverse.
Similarly,
if you evaluate a logarithmic
expression base A
at an exponential
of the same base,
you just get what's
in the power
of the exponential out,
and again that is found
from evaluating the function
at its inverse.
So let's take a look at this.
So log base 6
at 6 is equal to-- perfect, 1.
How about log base 12 of 12
to the 4th?
Do you see the bases match
so what do we get?
Awesome. We get out what's
in the power,
and you could have really done
this with the 6 to the 1.
What about over here?
Perfect. That's 0.
The power to which I raise 9
to get 1 is 1.
And this guy?
Good, you have an exponential
base 7 and a log base 7
so you get
out what's inside the log.
All right, moving along,
the product rule.
So if we now will have M, N,
and A, A is not equal to 1,
we make sure I wrote A not
equal-- I didn't even write it
up here, A not equal to 1.
All right, so M, N,
and A are positive
real numbers.
A is not equal to 1.
OK, so the logarithm
of a product turns
out to be the sum of the logs.
And we'll be proving this one
down below so you can see why.
So let's show why this
is true.
So proof, let's let X equal
to log base A at M
and let's let Y equal
to log base A at N.
So what happens
when we switch those two
exponential form?
Good. We would get--
working from this guy,
you see we would get A
because A is the base,
X is the power,
and M is what's inside
the log.
And so over here we'll get A
to the Y is equal to N. Now,
let's take a look
at log base A
of M times N. This is going
to be equivalent
to substituting an A to the X
for M, an A to the Y
for N. We'll have log base A
of A to the X times A
to the Y. So this is what we
substituted M.
[ Pause ]
All righty.
So, what is this
equivalent to?
Good. We have the same base
so we can add the exponents.
Now, using that property,
do you see we're evaluating
logarithm base A
at exponential base A,
so we're just going
to get X plus Y.
But back substituting,
what do we have?
X was-- Good, log base AM
and Y was log base AN.
So you proved it.
Awesome. OK.
So let's expand each
of these
logarithmic expressions.
If you want
to pause the movie, go ahead
and do so and then we will
like-- you can check
and see how you did.
All right,
let's see how you did.
This will be equivalent to--
we have a product in this log
so we'll have log base 6
at 6 plus log base 6
at X. Now, what can we do?
What is log--
the power to which we rate 6
to get 6?
Great, that's 1,
and then we still have log
base 6 of X here.
OK, so here, this is sort of--
I mean I'll show you
where this power rule is going
to be coming from soon.
Do you see X times X is
X squared.
But if we look at it
like this,
now this will be natural log
at X plus-- great--
natural log at X
which is how many natural log
of X's?
Awesome, two.
All right,
so now on to the
quotient rule.
Let's let M,
N and A be positive real
numbers and A is not equal
to 1.
So log base A of M divided
by N is equivalent
to log base A
at M minus log base A at N.
So the logarithm
as a quotient is the
difference of the logs.
So let's go ahead.
If you want
to try proving this one
yourself, go for it.
We'll show you why this
is true.
So starting off the same,
right, let X equal
to log base A at M
and let Y equal to-- perfect--
log base A at N.
So if we switched these
into exponential form,
we'll have A to the X is equal
to M and A to the Y is equal
to N. So log base A
at M divided
by N will be equal
to log base A at A
to the X divided
by A to the Y.
So again, A
to the X is substituted
in for M and A
to the Y is substituted
in for N. So this will give us
log base A of-- great--
A raised to the X minus Y.
And we'll have X minus Y
because we're evaluating a
logarithmic expression base A
at an exponential base A
so we get what's in the power.
And this was back substituting
log base A
at M minus log base A at N,
so you're done.
OK, so let's practice.
Expanding, we'll have log 1.
So what base does
that log have?
Good. It's the common
logarithm, so its base 1.
So that will be log 1 minus
log X but what is log 1?
Good, 0. So at the end
of the day,
we'll get negative log at X.
And over here,
we'll have log base 4
at X minus log base 4 at 2.
Can we do anything
with that log base 4 at 2?
Isn't the square root
of 4 equal to 2?
So the power
to which we raise 4 to get 2,
I believe, is?
Good, 1/2.
So typically what we'll do
here, notation-wise,
we'll bring that negative 1/2
to the front
so that you know it's not
in the log.
If I had meant for--
to write log at X minus 1/2,
I would have had parentheses
around the X minus 1/2.
OK, so next up,
the power rule.
So, let MN--
Actually, we don't need N
for this one.
So this is
where we get some edits.
Let M and A be positive real
numbers with A not equal to 1
and let R be any real number.
So what we'll have is log base
A at M raised
to the R is equal
to R times log base A at M.
So the logarithm
of a power is the product
of the power and the log,
and this is really powerful
and helpful
for solving
logarithmic equations.
So let's expand.
You remember what we did
in that earlier example,
natural log at X times X
and we got two
of those natural log of X's?
Well, now we have the
common logs.
So using that power rule,
the 2 is a multiplier
and we'll get 2 times log
at X. We're done.
Now, with this guy,
if you have a radical,
you may want
to write the radical using
rational exponents.
So now using the power rule,
we'll have 1/2 times log base
5 at X and we're done.
OK, so properties
for expanding
logarithmic expressions.
So basically this is just sort
of a summary of this stuff.
We want positive M,
positive N,
and then I'm just--
I want to keep the notation
the same.
Let's change these bases
to A. I don't want to switch
up the base and we're going
to change the R--
the P to an R, OK?
P is an R. And then of course,
we have A greater than 0
and A not equal to 1,
and we have R belongs
to the real numbers is an
element of the real numbers.
So, this is--
could kind of go
on the other direction.
If you see a sum of logs
and they have the same base,
you can condense it, right.
So, this will be log base A
at M times N. So, to expand,
right, you would write it
as a sum of logs.
To condense,
you would write it
as that product.
OK, so what's this guy going
to be?
Log base A at?
Perfect-- M over N.
If you want to do parentheses,
you can, you don't need to.
And then this one will be?
Awesome. Log base A
at M raised to the R.
So just a summary
of the properties we
just learned.
So let's expand these.
These are a little,
a little more mean.
So, what should we do?
We'll have-- perfect--
I'm thinking we should rewrite
this as log at X
to the 4th times Y minus 1
to the 1/3.
So then we have a product,
right, so we'll have log at X
to the 4th plus log
at Y minus 1 to the 1/3.
Using the power rule,
we have 4 times log
at X plus 1/3 times log
at Y minus 1, and you're done.
A lot of times people want
to do, you know,
make up a little
of their own math
with that log at Y minus 1.
But if you look
at the properties,
you'll see
that there would be no
property where you could bring
the log at--
to both the Y and the minus 1.
OK. How'd you do?
Awesome. Let's look
at this guy.
So let's just do this first
of all.
We have log base 2
and we have this whole
quantity that's being raised
to the 1/2.
So I'm going to move
over to the left a bit.
This will be 1/2 times log
at X squared plus 5 divided
by 12Y to the 6,
and then this will be 1/2
times log at X squared plus 5
and that's log base 2,
I apologize there.
I need to find my base,
and then minus log base 2
at 12Y to the 6th.
So then we'll get 1/2 times--
we're done
with the first expression
but the second one we can
write as log base 2
at 12 plus log base 2
at Y to the 6.
And then finally we'll have
1/2 times log base 2
at X squared plus 5 minus log
base 2 at 12.
There's technically a little
more we could do
with log base 2 at 12,
but we're good for this--
at this point for that.
And then we'll have minus 6
times log base 2 at Y,
and then we'll be all done.
[ Pause ]
All right.
Now, properties
for convincing.
This is when you're going
to go the other way
and I'm going
to make the same, you know,
changes that I had made
so that these look similar
to the properties we had done.
So, if we have for M greater
than 0 and N greater than 0,
A greater than 0, A not equal
to 1, and R is a real number.
So if you have a sum of logs,
you can condense it
and this is useful
for solving
logarithmic equations.
By the way,
what we were doing before is
really useful in calculus.
When you're doing derivatives,
it's much easier
to evaluate the derivative
of an expanded expression
than it is a
condensed expression.
OK, so here,
this will be log base A at M,
good, minus log base A at N,
and this one will be?
Perfect, R times log base A
at M. OK, so let's write these
as a single logarithm.
So you kind of go backwards,
if there's any powers you
bring them in.
So this one has--
we can rewrite the first term
as natural log at X cubed
and then minus natural log
at X minus 2 to the 1/4.
Now we have two logs
but have the same base of E
and we're subtracting them,
so we can write them
as one happy log
and they're going
to be divided, perfect.
And then if you want to,
you could rewrite this
so that you've got it
in a radical form.
Sorry, that base is kind
of big.
So the fourth root
of X minus two.
All right, let's do this.
Next guy, again,
please pause the movie,
give it a try yourself
and then check
and see how you did.
All right,
let's see how we did.
This will be equal to--
can't do anything
with the first term yet,
so let's bring that 12
in as a power, so it's going
to go N as a power.
Now, we have two terms,
both of them have a logarithm
at base four
and they're being added.
So, we can write this
as the product.
So, wait a minute.
You put base 4,
and it will be the product
of 5 times X plus Y raised
to the 12.
You're done.
All right, next up.
So this one again,
there's some weirdness here.
We're going to go ahead
and scribble this out.
Again, if you have the updated
edition, don't worry about it.
If you have the first edition,
you're lucky enough
to help discover the changes
with me.
So this proof--
or these two statements are
coming from the fact
that since Y equal
to log base A
at X is a
one-to-one function--
[ Pause ]
-- and if of course M is
greater than 0,
N is greater than 0--
[ Pause ]
-- A is greater than 0,
A is not equal to 1.
And this is how you're going
to solve--
this way you could solve some
log equations.
So basically, if M is equal
to N, then
if you can get a log equation
where you have the same base
log on either side, OK,
so this first property we
don't use as much in precalc
but you're allowed
to evaluate the log of--
Well, actually,
we use it a bunch.
I take that back.
If you have an unknown,
an unknown power, it--
like a variable in the power,
then what happens is you're
going to log both sides
of the equation
so that you can bring
that down as a base.
So this is
when you would log both sides
and we usually,
we'll natural log both sides
if for any base other than 10.
If you have a base of 10,
you would use the common log
of both sides so that's what
that property is for.
The second property here going
the other direction is
when you have a log equation
and it was expanded,
you do the work
to condense it,
then what you can do--
[ Pause ]
So if you have the same base
log on either side
and it's completely condensed,
then what's inside the logs
must equal each other.
OK? All right, so the change
of base property is another
powerful rule
so A is not equal to 1,
B is not equal to 1, and M,
they're all positive real
numbers, then log base A
at M is equal to log base B
at M divided by log base B
at A. Why would we use
this property?
Do you remember earlier
when I said, oh,
we can't quite check these,
you know, answers
yet because we don't know how
to graph logs with bases other
than 10 or E?
So here, this is going
to allow us to be able
to use our graphing calculator
to graph this and also
to evaluate them
on the calculator
because we only have the
natural log key
and the common log key.
So here we go,
use common logarithms
to evaluate log base 5 at 23
so log base 5 at 23 is equal
to log 23 over log 5
and that's approximately--
let's get our calculator out.
[ Pause ]
OK, so the log key is here,
so log 23 and make sure you
put the parenthesis
at the end before you do the
divided by and then log 5
and enter,
we get approximately 1.948.
All right, now,
let's use our natural logs,
so log base 5
at 23 equals natural log
at 23 divided by a natural log
at 5 which is approximately--
[ Pause ]
What do you know, 1.948.
What did you find out?
They're the same.
You get the same answer.
OK, so now, real quick,
what we might as well--
let's graph this one.
So if we go
to our graphing calculator
and we go to Y equals--
clear, clear--
So the way we would graph log
base 5 at 23 would be--
I'll just use the natural log.
Actually this one--
I'm sorry,
this one's just going
to be a constant,
let's do this.
Let's do-- because I think--
let's do this.
Let's graph log base 5
at X instead
and then we could plug 23
in to see--
and, you know,
to check an answer.
So this would be log X divided
by log 5, so natural log X
divided by natural log 5,
and look, that's our graph,
isn't that cool?
And then you could go
to second, calculate value
and put in 23, oops,
I need to expand my window.
[ Pause ]
And I think I'll just do Y max
of-- Well, yeah,
I think we'll be good.
All right, so now we'll go
to second-- calculate value.
Oops.
[ Pause ]
You know I think I did
something weird, OK.
Graph, third time's a charm,
right?
Second, calculate value 23,
enter, and look,
that's what we got, finally.
All right, here we go.
[ Pause ]
OK, so now you know how
to graph a log function
that has a base other than 10
or E, all right?
So, we might as well--
Here we go.
So, let's go
to our application.
So the application is using
this difference quotient
which you'll see a lot
in calculus and we want
to show that the difference
quotient for F at X equal
to log at X is this
expression here.
So let's do it.
F at X plus H minus F
at X divided
by H equals log base A.
So we'll replace X
with X plus H
and then the other part is
just log base A at X all
over H. So,
F at X plus H. Actually we'll
get down here, F at X plus H
with this.
And of course, F at X,
sometimes it's kind of like
so easy, it's hard, right?
F of X is what was given
up here, right?
OK, so this is going
to give us--
we have a difference
of two logs with base A,
so this is equivalent
to log base A
at X plus H divided by X,
divided by H. Well,
if we make a couple
of adjustments here,
if we divide through by X,
you see we'll get 1 plus H
over X, that's within the log,
and then we could times this
by 1 over H
when you're dividing
by H. It's the same
as timesing [phonetic] by 1
over H. And then let's just
switch the scalar
to the front, the constant
to the front and we have this.
And finally,
using the power rule
to condense, we'll have--
This guy will go
in as the power
and we'll have 1 plus H
over X raised to the 1 over H
and you're done, you did it.
Awesome. All right, so that,
I believe,
wraps it up for this lesson.
So, you have a fabulous day.
Bye.
------------------------------0e6dd376c6ed--
