### Video Transcript

In this video, weβll learn how to
perform operations and simplifications with expressions that involve rational
exponents.

You should already be familiar with
the laws of exponents for integer powers and πth roots. Recall that a rational number is
another name for a fraction. Both the numerator and denominator
are integers, that is, whole numbers, and π, the denominator, is nonzero. So, a rational exponent is simply a
power or exponent that is a fraction. So, if we have a number, a base π,
we can raise this to a rational power. If the numerator of our rational
exponent is equal to one, then we have our base π raised to the power one over
π.

And we say that a fractional
exponent one over π, where π is an integer, can be expressed as the πth root of
π. For example, eight raised to the
power one over three is the cubed root of eight. And thatβs equal to two since two
multiplied by two multiplied by two is equal to eight. That is, two is the number we cube
to get to eight. So, two is the cubed root of
eight.

Now, before we start looking at
some examples of simplifying exponential expressions with rational exponents, letβs
remind ourselves of the laws of exponents that apply to both rational and integer
exponents. We recall that we have two product
rules for exponents. The first is π raised to the power
π multiplied by π raised to the power π is equal to π raised to the power π
plus π. So, we sum our powers. And π raised to the power π
multiplied by π raised to the power π is π multiplied by π raised to the power
π. Our rule for negative exponents is
π raised to the power negative π is one over π to the power π.

Our quotient rules are π raised to
the power π over π raised to the power π is equal to π raised to the power π
minus π. So, we subtract the power. And π over π raised to the power
π is equal to π raised to the power π multiplied by π raised to the power
negative π.

Our power rules are π raised to
the power π to the power π is equal to π raised to the power π multiplied by
π. So, we multiply our powers. And π raised to the power π over
π, that is, a rational power or rational exponent, is equal to the πth root of π
raised to the power π. We note also that π raised to the
power zero is equal to one for any nonzero π. And one raised to the power π is
equal to one if π is a rational number.

So, letβs consider our first
example, where we use our knowledge of exponents to simplify an expression with a
rational exponent.

Simplify negative 64π raised to
the power 12 multiplied by π raised to the power 18 to the power one over six,
where π and π are positive constants.

Weβre given an expression raised to
a rational exponent. And to simplify this, weβre going
to make use of the power law for exponents, which says that π raised to the power
π to the power π is equal to π raised to the power π multiplied by π. First, however, we note that we can
decompose 64 into its prime factors. In fact, 64 is two raised to the
power six. So, now, separating out our
factors, we have negative two raised to the power six raised to the power one over
six multiplied by π raised to the power 12 to the power one over six multiplied by
π raised to the power 18 to the power one over six.

And now, using our power law, we
can write our expression as negative two raised to the power six multiplied by one
over six multiplied by π to the power 12 multiplied by one over six times π to the
power 18 multiplied by one over six. And since six multiplied by
one-sixth is equal to one, 12 multiplied by one-sixth is equal to two, and 18
multiplied by one-sixth is equal to three, we have negative two raised to the power
one multiplied by π squared multiplied by π cubed. And since any number to the power
one is itself, we have that the expression negative 64 multiplied by π to the power
12 multiplied by π raised to the power 18 all to the power one over six is equal to
negative two π squared π cubed.

Now, letβs try another slightly
more complex example, where we simplify an expression with a rational exponent.

Expand the negative of π cubed
plus one over π squared all raised to the power 10 all raised to the power one over
five, where π is a real constant.

Weβre given an expression in terms
of a real constant π, which is raised to the power 10. And we have the negative of this
expression, which is all raised to a rational power one over five. And to simplify this expression or
expand it, weβre going to use the power law for exponents, which tells us that π
raised to the power π to the power π is equal to π raised to the power π
multiplied by π. We note first, in fact, that our
negative sign means that weβre multiplying inside our parentheses by negative
one. And so, using our power law, we
have negative one raised to the power one over five multiplied by π cubed plus one
over π squared to the power 10 all raised to the power one over five.

Now, noting that multiplying five
instances of negative one together gives us negative one, then we can say that the
fifth root of negative one is also negative one. So now with this, together with our
power rule for exponents, we have negative π cubed plus one over π squared raised
to the power 10 multiplied by one over five. Our exponent 10 multiplied by one
over five is equal to 10 over five, and thatβs equal to two. And making some room, this leaves
us with negative π cubed plus one over π squared all squared.

And now, squaring our expression,
we have negative π raised to the power six plus two π cubed over π squared plus
one over π to the power four. Dividing the numerator and
denominator of our second term by π squared gives us two π. And multiplying through by negative
one, we have negative π raised to the power six minus two π minus one over π
raised to the power four. Hence, the expression negative π
cubed plus one over π squared to the power 10 all raised to the one over five is
equal to negative π raised to the power six minus two π minus one over π raised
to the power four.

Before our next example, letβs
recap again what we know about rational exponents. Fractional or rational exponents
are exponents with integer numerator and denominator, where the denominator π is
nonzero. And using the power rule for
exponents, we can say that π raised to the power π over π is equal to π raised
to the power one over π to the power π. And π raised to the power one over
π is the πth root of π. So, in fact, we have the πth root
of π raised to the power π.

Equivalently, we can take the
rational exponent on the outside. So, we have π raised to the power
π over π is equal to π raised to the power π all raised to the power one over
π. And thatβs equal to the πth root
of π raised to the power π. And so we have π raised to the
power π over π, the rational power, is equal to the πth root of π all to the
power π. And thatβs equal to the πth root
of π to the power π.

Weβll use this in our next example
together with prime factorization to break down our bases. This can help us to simplify some
complicated numerical expressions with different bases.

Simplify 36 to the power one over
four multiplied by 21 squared multiplied by eight raised to the power one over five
all divided by 486 raised to the power one over 10 multiplied by 42 cubed.

Weβre given a rational expression
to simplify, where the numerator and denominator comprise integers raised to
rational powers, that is, where the exponents are fractions. To do this, weβll use the laws of
exponents as shown. But first weβll decompose our
bases, thatβs 36, 21, eight, 486, and 42, into their prime factorizations. We know that 36 is three squared
multiplied by two squared, 21 is three times seven, eight is two cubed, 486 is two
multiplied by three raised to the power five, and 42 is two multiplied by three
multiplied by seven.

Our expression is now three squared
multiplied by two squared to the power one over four multiplied by three times seven
all squared multiplied by two cubed to the power one over five all divided by two
multiplied by three to the power five all to the power one over 10 multiplied by two
times three times seven all cubed. Using our rule number one, we can
bring our exponents inside our parentheses. And we can also use rule four at
the same time; thatβs the power rule. And noting that two over four is
equal to one-half and also that five over 10 is equal to one-half, and collecting
our like bases in both numerator and denominator, we have three raised to the power
of a half multiplied by three squared multiplied by two raised to the power of a
half multiplied by two to the power of three over five multiplied by seven squared
divided by three raised to the power of a half multiplied by three cubed multiplied
by two raised to the power one over 10 multiplied by two cubed multiplied by seven
cubed.

And at this point, we can use our
power rule three and our negative exponent rule number five to collect our exponents
for each of our bases so that we have three raised to the power of a half plus two
minus one-half minus three multiplied by two raised to the power of a half plus
three-fifths minus one-tenth minus three multiplied by seven raised to the power two
minus three. Evaluating one-half plus two minus
one-half minus three, this gives us negative one. And similarly, for our power of
two, we have one-half plus three-fifths minus one-tenth minus three is equal to
negative two. And two minus three for seven is
negative one. And so we have three raised to the
power negative one multiplied by two to the power negative two multiplied by seven
raised to the power negative one.

And again, using our rule number
five, we have one over three multiplied by two squared multiplied by seven. And evaluating this gives us one
over 84. And hence, our expression 36 to the
power one over four multiplied by 21 squared multiplied by eight to the power one
over five all divided by 486 to the power one over 10 multiplied by 42 cubed
simplifies to one over 84.

In the case where our bases are
decimal, we would write this base as a fraction and then decompose the numerator and
denominator of this fraction into their prime factors. So, letβs look at an example of
this.

Simplify 0.25 raised to the power
three over two multiplied by 1.8 squared all over eight raised to the power two over
three.

In this example, we want to
simplify a numerical expression which combines decimals and rational exponents. Letβs begin by noting some of the
laws of exponents that weβll need. Before we use these laws, however,
letβs write each of the decimal bases as a fraction. We can then express these in terms
of their prime factorizations. Our bases are 0.25, 1.8, and
eight. And we know that 0.25 is
one-quarter, and thatβs equal to one over two squared. 1.8 is equal to nine over five, and
thatβs three squared over five. And we know that eight is equal to
two cubed. Our expression can then be written
as one over two squared raised to the power three over two multiplied by three
squared over five all squared all over two cubed raised to the power two over
three.

Now, using rules two and rule four,
our expression becomes one over two raised to the power two multiplied by three over
two all multiplied by three to the power two times two all over five squared. And thatβs all over two raised to
the power three multiplied by two over three. And evaluating our exponents, we
have two multiplied by three over two is equal to three, two squared is four, and
three multiplied by two over three is equal to two. What we now have then is three to
the power four divided by two cubed multiplied by two squared multiplied by five
squared. We can sum our powers of two using
rule number three so that we have three raised to the power four divided by two to
the power five multiplied by five squared. And evaluating this gives us 81
over 800. Our expression then 0.25 raised to
the power three over two multiplied by 1.8 squared all divided by eight to the power
two over three simplifies to 81 over 800.

In our final example, we simplify
an expression involving rational and negative variable exponents.

Simplify 25 raised to the power
three over two π₯ multiplied by eight raised to the power π₯ minus five over three
all divided by 100 raised to the power three over two π₯ multiplied by the square
root of 100.

The expression we want to simplify
in this example involves rational and negative exponents. So, letβs make a note of the laws
of exponents weβre going to need to use. Now, assuming that π₯ is a real
number, letβs begin by looking at the exponents in each of the factors. In our first factor, thatβs 25
raised to the power three over two π₯. We can use rule number four to
bring in the power of a half. And since we know that 25 raised to
the power of a half is the square root of 25, and thatβs equal to five, our first
factor is five raised to the power three π₯.

Our second factor in the numerator
is eight raised to the power π₯ minus five over three. And we can use the product rule
number three to break this down to eight to the power of π₯ multiplied by eight to
the power negative five over three. Next, using rule number four on the
second part of this, we have eight raised to the power π₯ multiplied by eight to the
power one over three to the power negative five. Now we know that eight raised to
the power one over three is equal to two because eight to the one over three is the
cubed root of eight. And since eight is equal to two
multiplied by two times two, thatβs equal to two cubed. The cubed root of eight is two. So, in fact, eight raised to the
power π₯ minus five over three is eight to the power π₯ multiplied by two to the
power negative five.

And now using rule number four
again on 100 to the power three over two π₯, we have 100 to the power of a half to
the power three π₯. And we know that 100 to the power
of a half, that is, the square root of 100, is equal to 10. So that 100 to the power three over
two π₯ is equal to 10 to the power three π₯. So, now making some room, we have
25 raised to the power three over two π₯ is equal to five to the power three π₯. Eight to the power π₯ minus five
over three is eight to the power π₯ multiplied by two to the negative five. And 100 to the power three over two
π₯ is 10 raised to the power three π₯. And our final factor in the
denominator is the square root of 100, which is 10.

Our expression now becomes five
raised to the power three π₯ multiplied by eight to the power π₯ multiplied by two
to the negative five all over 10 to the power three π₯ multiplied by 10. So, now, decomposing our bases of
eight and 10 into prime factors, we know that eight is equal to two cubed and 10 is
two times five. So, now, we have five to the power
three π₯ multiplied by two cubed to the power π₯ multiplied by two to the negative
five power all over two times five to the power three π₯ multiplied by two times
five.

And now using rules one and four,
we can bring our powers in. We can divide both our numerator
and denominator by five raised to the power three π₯ and also by two raised to the
power three π₯. And weβre left with two raised to
the power negative five divided by two multiplied by five. Using rule number five, we see the
two to the negative five is one over two to the power five. And so we have one over two
multiplied by two to the power five multiplied by five. And by rule number three, this is
one over two to the power six multiplied by five. And since two to the power six is
64, we have one over 64 multiplied by five, which is one over 320.

The expression 25 raised to the
power three over two π₯ multiplied by eight to the power π₯ minus five over three
all divided by 100 to the power three over two π₯ multiplied by the square root of
100 simplifies to one over 320.

Now, letβs complete this video by
recapping some of the key points weβve covered. Weβve used the laws of exponents
for the bases π and π and rational exponents π, π. We have the product rules π raised
to the power π multiplied by π to the power π is equal to π raised to the power
π plus π; thatβs the sum of the powers. π to the power π multiplied by π
to the power π is π multiplied by π all to the power π. We have the negative exponent rule,
which is π raised to the power negative π is one over π to the power π. Our quotient rules tell us that π
raised to the power π over π to the power π is equal to π to the power π minus
π. And π to the power π over π to
the power π is π over π all to the power π.

And finally, our power rules π to
the power π all to the power π is π to the power π multiplied by π. And π to the rational power π
over π is the πth root of π raised to the power π. We note also that π raised to the
power zero is equal to one for any π not equal to zero and that one raised to the
power π is equal to one for a rational number π. When simplifying numerical
expressions with different bases or exponents, we can use prime factorization to
simplify the bases and collect like bases. We then use the laws of exponents
to simplify our expressions.