Hey, guys.
Welcome to Algebra 1.
Today's lesson focuses
on negative exponents.
Your knowledge on the laws
of exponents
and quotients and products
will get you through
this lesson.
Ready? Let's go.
Okay.
Think about this problem
in two separate parts.
First thing I'll do
for the right side of it
is expand it out.
X cubed divided
by X to the 5th,
I know for my numerator
that means
that I have X times X
times X.
When raising X
to the 3rd power,
I'm just multiplying
by three Xs.
For the denominator,
if I have X to the 5th,
then that means X times X
times X
times X times X.
Let's make
that fraction bar longer.
For the denominator,
with X to the 5th power,
that means I'm multiplying
five Xs together, okay?
Now, knowing what you do about
simplifying fractions,
if you have
the same thing
in the numerator
and the denominator,
then cancel things out.
I can cancel out
one X up top
with one X in the bottom,
then another on top
with another on the bottom.
I'll cancel out another
on the top and bottom.
Now here's a little trick.
We don't have a value of 0
on top,
but you know there's always
that invisible factor of 1?
I'm going to throw up here
that this is like X times X,
times X, times 1.
Just throw that out there.
I have canceled out
these three Xs up top
with these three X's
on the bottom,
and what I'm left with
is a 1 up top
and I'm left with
X times X on the bottom.
Right?
Now, I could represent that
differently.
I could say, well,
1 over X times X,
that's really the same thing
as 1 over X squared.
Right? Okay.
That's how
the right side looks.
I'll box that in
to focus on that for the right.
Now go back to the left side,
how it started.
We had X cubed
divided by X to the 5th.
One of the laws of exponents
is, with a
setup like this,
when we're dividing powers,
we can subtract the exponents.
I could represent this
as X to the 3 minus 5,
because I need to subtract those
exponents to simplify them.
What is 3 minus 5?
Let's come off to the side.
3 minus 5,
that's negative 2.
Knowing what you do
about integer operations--
or just put that
in the calculator if want.
X to the 3 minus 5,
that's the same thing
as X to the negative 2.
What that means is X
to the negative 2nd power
is really the same thing
as 1 over X to the 2nd power.
Sounds a little crazy
at first.
Let's do a couple more
to make it make sense.
Let's take this one.
Going to apply
that same process.
So X to the 4th, I know
that's the product of four Xs:
X times X times X times X.
Okay?
Then on my denominator,
X to the 7th,
that's the product
of seven Xs.
So X times X times X
times X
times X
times X times X.
Let's count that.
One, two, three, four,
five, six, seven--yes.
I'm going to expand,
draw that fraction line
longer.
On the last one,
I added that factor of 1,
because multiplying by 1
doesn't change anything.
That's how I'll start.
So X times X times X
times X,
I'll multiply times 1--
throw that in there, okay?
Didn't change
the problem at all.
Now, let's start
canceling things out.
I cancel out one X
in the numerator
with one in the denominator;
cross them out.
I cancel out another X
in the numerator
with one
in the denominator;
one in the numerator
with one in the denominator;
and another one
in both places.
And what I'm left with
is a 1 in the numerator
and then the product
of three Xs in my denominator.
So I'm left with 1 over X
times X times X.
I'm going to represent that
differently,
and say, well,
this is the same thing
as 1 over X to the 3rd power,
right?
So we expanded it out
and looked at it that way.
Let's look
at how this problem started,
this X to the 4th
over X to the 7th.
Knowing the laws of exponents,
when you have to find
the quotient of powers,
subtract your exponents.
I have X to the 4th,
divided by X to the 7th,
so that's X to the 4 minus 7,
right?
So 4 minus 7--
come off to the side--
that's negative 3.
I could represent that
as X to the negative 3.
So X to the negative 3rd power
means the same thing
as 1 over X
to the positive 3rd power.
Are you starting to notice
the pattern?
Let's try another one.
X cubed
over X to the 8th.
I know I can represent X cubed
as X times X times X,
and I'll throw in
that times 1,
because it didn't
change anything.
I have the product
of eight Xs on the bottom,
so X times X
times X times X
times X times X
times X times X.
Let's double check: three, four,
five, six, seven, eight--
we do have eight Xs up there.
I'll start
canceling things out,
let's start simplifying.
I cancel out one X each
in the numerator
and the denominator;
another one
in the numerator
with one
in the denominator;
and another in the numerator
with one
in the denominator.
When you do that, you're left
with that 1 in the numerator,
and you have one,
two, three, four,
five Xs
in your denominator.
You can simplify that,
write that
a little differently.
I'll leave my 1
in the numerator.
And since I'm multiplying
X by itself five times
in my denominator,
then I can represent that
as X to the 5th, okay?
We handled the right side;
now let's handle the left side.
X to the 3rd
divided by X to the 8th,
apply that law of exponents.
For the quotient of powers,
you know you subtract
your exponents,
so I need to simplify that
as X to the 3 minus 8.
Go ahead and clean that up,
come off to the side.
What is 3 minus 8?
That's negative 5.
So that means
that X to the 3 minus 8
simplifies to X
to the negative 5th power.
So X to the negative 5th
means the same thing
as 1 over X to the 5th.
Did you see that pattern?
Each and every time,
if you notice,
when you raise a value
to a negative power,
it's the same thing
as 1 over that value
to the positive power.
That's how you handle
negative exponents.
Let's look at this rule.
Any value raised
to the negative power--
any negative power--
it's the same thing
as 1 over raising that value
to a positive power,
except for one value: 0.
Let's get a pen.
When you raise 0
to a negative power,
this gets a little heavier
than Algebra 1,
so you won't really see this
too much in this course.
You can just focus
on this rule,
knowing that A
will never be 0.
Any value, except 0
raised to a negative power,
means the same thing
as 1 over that value
to a positive power,
that same power but positive.
Let's try a couple problems.
We have 5X and we raise that
to the negative 2nd power.
First,
since I know my rule
that anything raised to a
negative power means the same
as 1 over raising that same
value to a positive power,
I'm going to rewrite this
as 1 over 5X
to the positive 2nd power,
because those values mean
exactly the same thing.
I'll work
with that positive exponent
to help me make
my problem simpler.
I'm going to keep
simplifying this.
In my denominator,
I have a product
and I'm raising it
to a power,
so I apply
that rule of exponents.
On the side,
I'll do a little work.
I raise that 5
to the 2nd power
and that X
to the 2nd power.
We'll put our work here.
We'll raise our 5
to the 2nd power
and that X to the 2nd power.
Let's keep chugging along.
5 squared,
I know that's 25,
and X squared,
it kind of took care of itself;
you got X squared.
Once you're there,
then you're all done.
You can't take it any further.
So 5X raised
to the negative 2nd power
means the same thing
as 1 over 25X squared.
Let's apply this again,
give you a little practice here.
We have X to the 5th,
Y to the 8th over X squared,
Y to the 10th.
This one's a little different,
but remember,
don't memorize my steps,
memorize my process--
that's what you want to learn.
I have a quotient
and I have a product up top.
I'm dividing it
by a product on the bottom.
I have
some of the same bases:
an X and Y on top,
X and Y on the bottom.
That's my cue to group
the like things together.
I'm going to handle
the Xs first.
I need X to the 5th
over X squared,
and then I'll multiply
that by Y to the 8th
over Y to the 10th.
I broke it up to help me
deal with it.
Knowing what I do about
the laws of exponents,
with a quotient of powers,
I must subtract the exponents
to make it simple.
Here, I have a 5 and a 2,
so it's 5 minus 2--
come off to the side
for that--5 minus 2, that's 3.
So as far as my Xs go,
I have X to the 3rd power.
For my Ys,
I have an 8 and a 10;
I need to subtract those.
So 8 minus 10,
that's negative 2.
Our negative exponent
popped up in this problem.
So Y
to the negative 2nd power.
I'm going to keep going
across with this one.
X cubed, I don't have
a negative exponent,
I don't need
to do anything.
I'll come underneath.
For the negative exponent,
for Y
to the negative 2nd power,
I know I can rewrite that
as 1 over Y
to the positive 2nd power.
Those values mean
exactly the same thing.
In algebra, you don't want
negative exponents
as your answer.
When you've taken it
as far as you can go,
your exponents
should be all positive.
That's why we kept going.
You could stop here and you
wouldn't be wrong necessarily,
but you'll see it
written like this.
Go ahead and multiply
those two terms together,
and you'll see it written
as X cubed over Y squared.
That's how you'll see
that answer written--
we multiplied
our numerators together.
X cubed times 1,
that's just X cubed.
And you remember fractions,
you can always make
something whole a fraction
by putting it over 1.
1 times Y squared,
that's Y squared.
That's how the problem
went from this to this.
But either one
is really acceptable.
Okay?
Let's keep going.
We have C to the 8th,
D to the negative 5th over 3
and we're raising that entire
quotient to the 3rd power.
We apply a couple
of laws of exponents here.
I'll start
by breaking this up,
separating my numerator
from my denominator.
I'm going to write this
as C to the 8th,
D to the negative 5th,
and I'm raising that
to the 3rd power,
over 3 to the 3rd.
I just
distributed that exponent
throughout those pieces
in my quotient.
Now that I see the numerator,
it's a product,
I need to keep on carrying
that exponent throughout.
I need to raise C to the 8th
to the 3rd power,
so C to the eighth
to the 3rd.
And I need to raise D to the
negative 5th to the 3rd power,
so D to the negative 5th
to the 3rd power.
For my denominator,
I have 3 to the 3rd.
I can take care of that:
3 to the 3rd, that's 27.
So the denominator
is done;
we've got to clean up
the numerator here.
So you know about raising
a power to a power, right?
Here you need to multiply
your exponents together.
So 8 times 3, that's 24.
So that first part
became C to the 24th,
and negative 5 times 3,
that's negative 15,
so D to the negative 15th.
And then we'd already
taken care of that 27.
Not quite done yet because
we have that negative exponent
so we've got to keep
cleaning this up.
Let me shift a little work
around here,
get some more space.
Since we've taken
this problem from this step
to this step, I'll
get rid of this up top
to focus on
where we are right now.
I think I'll also
get rid of that.
Now, oop! I do want to see
my work here.
Let's bring this work up top.
Let's get rid of
that slash-through,
because we didn't want
to slash through that.
Let's keep going.
Get the pen back,
here we go. All right.
We need to handle that negative
exponent, that negative 15.
Now, here's when
you can use a shortcut.
The rule tells us,
if I rewrite the rule--
if we have X to the negative A
power--I'll change it up--
that's the same as 1 over X
to the positive A power.
As a shortcut, you can
think about this is,
if your negative exponent
starts out on top
in a quotient,
drop it to the bottom
and make it positive.
Had the negative exponent
been at the bottom,
we'd need to bring it up top
and make it positive.
It's like you're shifting
its position, like a shortcut.
What that means is I can
rewrite this as C to the 24th.
Because this part
of the term
has a negative exponent,
I'll bring it to the bottom.
I still have that 27,
so I'll drop that D
to the bottom, changing that
exponent to a positive 15.
You've applied that same rule
for negative exponents;
you just took a shortcut
to get there.
That's the trick with algebra:
once you see patterns,
algebra is really
a bunch of shortcuts
that people figured out
over time by noticing patterns.
Use your knowledge of patterns
to really help you
get through the course.
It is your turn to try.
Look at these three problems,
pause the tape,
take a few minutes
and work through them.
When you're ready
to compare answers,
press play.

(female narrator)
Simplify.
Number 1:
6Y to the negative 2nd power.
Number 2: A to the 7th,
B to the 6th
over A to the 11th, B.
Number 3: S to the negative
3rd, T to the 4th, over nine,
all to the 2nd power.
You ready to get going?
Let's compare our answers.
For the first problem,
you had 6Y
and you were raising it
to the negative 2nd power.
You should have got 1
over 36Y squared for that one.
Number 2,
you had A to the 7th,
B to the 6th,
over A to the 11th, B.
That simplified
to B to the 5th,
A to the 4th--
over A to the 4th.
Number 3, you had
S to the negative 3rd,
T to the 4th, over 9,
and you were squaring
that quotient.
That answer was T to the 8th,
81S to the 6th.
To see how I did these,
stay with me.
Let's look
at that first one.
I have 6Y;
I'm raising that product
to the negative 2nd power.
I'll apply my rule
about negative exponents
which tells me that I can
represent this
as 1 over 6Y
to the positive 2nd power.
Now I just need
to simplify this denominator.
I've got my 1 on top
and that's good.
I'll raise
these individual pieces
of this product
to the 2nd power.
I have 6 to the 2nd power
times Y to the 2nd power.
Then just simplify this
a little further.
The 1 is good to go up top;
6 squared, that's 36.
And Y to the 2nd power
took care of itself--
it's already simple.
That's how I got
that first answer,
that 1 over 36Y squared.
Need to see the second one,
here we go.
I had A to the 7th,
B to the 6th,
over A to the 11th, B.
First,
I broke this up.
I grouped the like things
together.
I looked at it as A to the 7th
over A to the 11th,
and I was multiplying that
by B to the 6th over B.
Then I used
the laws of exponents,
to know that if I have
a quotient of powers,
then I need to subtract
the exponents.
So for the As,
you have 7 and 11.
Come off to the side here--
7 minus 11, that's negative 4.
The first part of that
was A to the negative 4th.
For the Bs, I had B
to the 6th, then B.
When you don't
see an exponent,
there's an invisible
exponent of 1.
Always remember,
if you don't see an exponent
there's a 1 there.
Just write it in.
For this one, B to the 6th
over B to the 1st.
Subtract your exponents here,
so 6 minus 1, that's 5.
I know this was
times B to the 5th
and knew I wasn't done because
I had a negative exponent.
Use what we know
about patterns
and take a shortcut here.
That A to the 4th is part
of the numerator now.
I need to drop it to the bottom
and make it positive.
That is how I got
that second answer.
If you need to see
the third one, here we go.
We had S to the negative 3rd,
T to the 4th, over 9.
We were squaring
that whole quotient.
I started out
by breaking it up.
I took my numerator,
S to the negative 3,
T to the 4th squared,
over 9 squared.
I used what I knew about
exponents to apply the rule
of raising a product
to a power,
so I raised
S to the negative 3rd power,
to the 2nd power.
I needed to take
these pieces separately.
Now I need to raise T
to the 4th to the 2nd power,
so T to the 4th
to the 2nd power,
so I handled the numerator.
Now, for the denominator,
9 squared,
that's 9 times 9,
so that's 81.
My denominator's
all taken care of,
but I'm not done yet.
I've got to handle
that numerator.
So raising a power to a power,
negative 3 times 2,
that's negative 6.
So S to the negative 6.
4 times 2 here, that's 8,
so T to the 8th.
Then that was all over 81.
You have
a negative exponent,
so you're not done yet.
I may be able
to fit that over here.
We use what we know
about negative exponents
for a shortcut.
T to the 8th isn't
going anywhere, it's fine.
That S to the negative 6--
I'll take care of that.
Apply the rule
about negative exponents.
81 is good.
I'll drop that S
to the bottom,
making that exponent positive.
Because remember,
the rule tells us
if I'm raising something
to a negative power,
it means the same thing
as 1 over that
to a positive power.
You can follow
that pattern
and know you need to change
its position in the fraction.
We were all done there.
I hope you're feeling
more comfortable
with negative exponents,
that you saw
how products and quotients
and factors--
all that knowledge
helped you get through
these problems.
See you soon. Bye!