### Video Transcript

In this video, we will learn how to
use the rules of negative and fractional indices to solve algebraic problems. In order to help us understand
these rules, letβs begin by recalling the rules for multiplication and division of
exponents. Here we have the rules for
multiplication and division of exponents. For values of π and π in the real
numbers, then we can multiply exponents with the same base, and we do that by adding
their powers. Then, to divide exponents that have
the same base, we subtract their powers. This time, π has to be a nonzero
value. Now, since π and π can be any
real values, then these rules will also apply for negative and fractional
indices. Letβs consider what happens when we
modify these rules so that we obtain a negative exponent.

Using the division law, letβs take
the value of π to be equal to zero. Then, we would have π to the power
of zero divided by π to the power π. Using this exponent rule, we can
write this as π to the power of zero minus π. This is equivalent to π to the
power negative π. So letβs consider what we have just
discovered by using the fact that π to the power zero equals one. Itβs the fact that one divided by
π to the power π or one over π to the power π is equal to π to the power of
negative π. We can add this to the rules of
exponents as a rule for negative indices. Notice that here π and π are
still values in the real numbers and π is nonzero.

In the first example, we will see
how we can apply this exponent rule.

Which of the following is equal to
negative 10 over nine π₯ to the power negative two π¦ to the power negative
seven? Option (A) negative nine over 10π₯
squared π¦ to the power seven. Option (B) negative 10 over nine π₯
to the power seven π¦ squared. Option (C) negative 10 over nine π₯
squared π¦ to the power seven. Or option (D) negative 10π₯ squared
π¦ to the power seven over nine.

In this question, we have some
negative exponents. And so it would be useful to recall
the law of exponents for negative exponents. This law states that π to the
power negative π is equal to one over π to the power π for any π which is
nonzero. Because both π₯ and π¦ have
negative exponents here, then we can apply the rule to both variables. For π₯ to the power negative two,
we can substitute π is equal to π₯ and π is equal to two. So π₯ to the power of negative two
is equal to one over π₯ squared. In the same way, for π¦ to the
power negative seven, that means that π is equal to π¦ and π is equal to
seven. So π¦ to the power negative seven
is equal to one over π¦ to the power seven.

We can then substitute these values
into the expression. This gives us negative 10 over nine
times one over π₯ squared times one over π¦ to the power seven. And when weβre multiplying
fractions, we multiply the numerators and multiply the denominators. The expression is therefore equal
to negative 10 over nine π₯ squared π¦ to the power seven, which was the answer
given in option (C).

In the next example, we will use
the exponent rule for negative indices as well as the rule of division for
indices.

True or False: The simplified form
of π₯ to the power negative four over π₯ to the power negative two is one over π₯
squared.

One way in which we can simplify
this expression is by using the division rule for exponents. This rule tells us that if we are
dividing two exponents, in this case π to the power π divided by π to the power
π, then we subtract those indices such that we get π to the power π minus π for
any π which is nonzero. And we know that this value of π₯
to the power negative four or π₯ to the negative fourth power over π₯ to the power
negative two is equal to π₯ to the power negative four divided by π₯ to the power
negative two. Using the exponent rule with π
equal to negative four and π equal to negative two, we would have π₯ to the power
of negative four minus negative two. This simplifies to π₯ to the power
negative two.

This is a perfectly valid
equivalence, but it doesnβt match the form that we were given in the question. However, we should remember that
there is another law of exponents for negative exponents, which states that π to
the power of negative π is equal to one over π to the power π for a nonzero value
of π. So this value of π₯ to the power of
negative two is actually equal to one over π₯ squared. Thatβs using a value of π equal to
two. So what we have done here is work
out that this expression π₯ to the power negative four over π₯ to the power of
negative two is indeed equivalent to one over π₯ squared. And so the statement in the
question is true.

But thereβs also another way in
which we couldβve manipulated this expression to come to the same conclusion. In this alternative method, we
would actually start by applying this rule for negative indices first. When we do this and we take the
numerator of this expression first, we can say that this is equivalent to one over
π₯ to the power four. The denominator of π₯ to the power
negative two is equivalent to one over π₯ squared. Simplifying this is easier if we
remember that a fraction is the same as a division. So this would be equal to one over
π₯ to the fourth power divided by one over π₯ to the second power.

We divide two fractions by
multiplying by the reciprocal of the second fraction. So we are left with π₯ squared over
π₯ to the power four. In order to simplify this even
further, we could apply the first rule that we saw in this question. With a value of π equal to two and
π equal to four, we would have π₯ to the power of two minus four. And of course two minus four is
negative two. And then we can rewrite this using
our rule for negative indices as one over π₯ squared. And thus, we have confirmed that
the statement in the question is true.

Before we look at any more
examples, letβs recap some more rules of exponents. These exponent rules for powers
tell us that for any π and π in the set of real numbers, we have that π to the
power π to the power π is equal to π to the power ππ. ππ to the power π is equal to π
to the power π times π to the power π. And thirdly, π over π to the
power π is equal to π to the power π over π to the power π, where π is
nonzero.

Letβs see how we can apply these
exponent rules in the following examples.

Simplify π over π to the power
negative one all to the power negative three times two π to the power negative two
over π to the power negative two all to the power negative three.

In order to simplify this
expression, we will need some of the rules of exponents. Because we have fractions raised to
a power, then we can use one of the power laws, which tells us that π over π to
the power π is equal to π to the power π over π to the power π, where π is
nonzero and π is in the real numbers. So letβs apply this rule to the
first part of the expression. As we have this fraction to the
power negative three, then we know that this will be equivalent to the numerator to
the power negative three over a denominator to the power negative three. However, to simplify the
denominator of π to the power negative one to the power negative three, weβll need
another rule of exponents.

The rule that we need is one of the
power laws, which tells us that π to the power π to the power π is equal to π to
the power of π times π. We take the two exponents of
negative one and negative three and multiply them. And we know that negative one
multiplied by negative three is three. We have now simplified this part of
the expression to π to the power negative three over π to the power three. Letβs see if we can simplify the
second part of this expression in the same way.

The first thing we can do is apply
this rule for powers of fractions. So the numerator will be equivalent
to two π to the power negative two to the power negative three. And the denominator will be π to
the power negative two to the power negative three. In order to simplify the numerator
of this fraction, we will need to recall another law of exponents. This power law tells us that ππ
to the power π is equal to π to the power π times π to the power π, where π is
in the set of real numbers. The numerator of this fraction will
therefore simplify to two to the power of negative three times π to the power of
negative two to the power of negative three. We can also simplify the
denominator, remembering that we can use this second power law here to multiply the
exponents. And negative two times negative
three gives us six, so the denominator will become π to the power six.

The next stage in our working will
be to simplify this part of the expression, π to the power of negative two to the
power negative three. Just as before, we can multiply
these exponents. So we have π to the power of
negative two times negative three. And we know that negative two times
negative three is six. Now, we could potentially simplify
this expression a little further by dealing with the two to the power of negative
three. But for now, letβs substitute these
values in orange and pink for the parts of the expression. When we multiply these together, we
have π to the power negative three over π to the power three times two to the
power negative three π to the power six over π to the power six.

We know that when we multiply
fractions, we multiply the numerators and multiply the denominators. We might then notice that on the
numerator, we have two values of the same base of π. And there is an exponent rule to
help us work this out. This rule tells us that π to the
power π times π to the power π is equal to π to the power of π plus π. Of course, the values that weβre
using in the question of π and π are not the same values that weβre using in these
exponent rules. And so on the numerator, we will
add the two exponents for π of negative three and six. So we have two to the power
negative three times π to the power of negative three plus six on the
numerator. On the denominator, we add the
exponents three and six of π. So we have π to the power of three
plus six.

At this point, we have simplified
the variables of π and π as much as we can. But letβs see if we can do anything
to simplify this two to the power of negative three. And we can use one final exponent
rule for negative indices. This rule tells us that π to the
power of negative π is equal to one over π to the power π. This means that two to the power of
negative three can be written as one over two cubed. We should remember that two cubed
is equal to two times two times two, and thatβs eight. So two to the power of negative
three is equal to one over eight. And when we plug in one-eighth in
place of two to the power negative three, we have the expression π cubed over eight
π to the power nine. And this is the answer. We have simplified the given
expression as much as we can to give π cubed over eight π to the power nine.

Letβs now see one final
example.

Simplify the expression π₯ to the
power eight over π¦ to the power negative four all to the power one-half.

In order to simplify this
expression, we can start with the power law for fractions, which tells us that π
over π to the power π is equal to π to the power π over π to the power π. We can therefore write that this
expression is equal to π₯ to the power eight to the power one-half over π¦ to the
power negative four to the power one-half. In order to simplify the powers on
the numerator and denominator, we can apply a second exponent rule. We multiply the exponents on the
numerator of eight and one-half and the exponents on the denominator of negative
four and one-half. This gives us π₯ to the power four
over π¦ to the power negative two. While this is a perfectly valid and
fully simplified expression, it is commonplace to give negative exponents instead as
a positive exponent.

We recall that π to the power of
negative π is equal to one over π to the power of π, where π is nonzero. This means that we can write our
expression as π₯ to the power of four over one over π¦ squared. We can simplify this fraction
within a fraction by remembering that fractions are all about division. What we actually have here is the
expression π₯ to the power of four divided by one over π¦ squared. We remember that to perform a
division by a fraction, we multiply instead by its reciprocal. That means we have π₯ to the power
of four multiplied by π¦ squared. And so the answer is that the
expression given in the question can be simplified to π₯ to the power of four π¦
squared.

Before we finish this video, there
are two more rules of exponents that we need to know. These are regarding fractional
indices. The first rule tells us that π to
the power of one over π is equal to the πth root of π for any value of π greater
than or equal to zero and any positive integer π. We can also extend this to give us
a second rule that π to the power of π over π is equal to the πth root of π to
the power π. And thatβs also equivalent to the
πth root of π to the power π. These two rules are particularly
useful when weβre using numerical values rather than algebraic values.

But now letβs summarize the key
points of this video. We saw that the laws for
multiplication, division, and powers of indices apply also for fractional and
negative indices. We then saw that the law of
exponents for negative indices is one over π to the power π is equal to π to the
power of negative π for nonzero values of π. And finally, we finished with two
laws of exponents for fractional indices. It is worth noting all of these
exponent laws as we are often required to learn them for examinations.