### Video Transcript

In this video, weβre going to
simplify expressions using rules of exponents.

The first question that comes up
is, what is simplifying expressions? By simplifying, we mean writing an
expression in the most compact or efficient manner without changing the value of the
expression. For this process, we might do
things like removing parentheses or brackets by multiplying factors, combining like
terms, and, as weβll look at today, simplifying exponents. And why would we use this kind of
process?

Simplifying expressions makes them
easier to read and understand. It also reduces possible errors in
calculating, as after an expression is simplified, there are less calculations that
will need to occur. Letβs say, we have an exponent in
the form π times π₯ to the π power. In this case, the constant in front
of the π₯ will be the coefficient. Whatever is being taken to a
certain power will be the base of the exponent. The superscript in the upper
right-hand corner of the base is called the power or the index. And when all of this together is
inside an expression, it will be one of the terms in the expression.

Now, weβre ready to consider some
rules about how we operate with exponents. Hereβs the first rule: π₯ to the π
power times π₯ to the π power equals π₯ to the π plus π power. We might see something like this:
π₯ squared times π₯ cubed. Weβre able to simplify that by
writing π₯ to the two plus three power, π₯ to the fifth power. Another rule to consider is π₯ to
the π power over π₯ to the π power which is equal to π₯ to the π minus π
power. If we had something like π₯ to the
sixth power over π₯ to the fourth power, we could simplify that by saying π₯ to the
six minus four power, or simply π₯ squared.

The next rule we call a power of a
power. π₯ to the π power to the π power
is equal to π₯ to the π times π power. For example, π₯ to the seventh
power to the third power equals π₯ to the seven times three power, or simply π₯ to
the 21st power. Before we move on, with this rule,
we can also note that π₯ cubed to the seventh power will also be equal to π₯ to the
three times seven power, which is π₯ to the 21st power. And that means π₯ to the seventh
power cubed is equal to π₯ cubed to the seventh power.

Next, we have π₯ times π¦ to the π
power will be equal to π₯ to the π power times π¦ to the π power. π₯π¦ cubed is equal to π₯ cubed π¦
cubed. We should also say that this rule
can be combined with a power to a power rule. If we have π₯ squared times π¦ to
the fourth to the third power, weβll have π₯ to the two times three power times π¦
to the four times three power, π₯ to the sixth power and π¦ to the 12th power.

The same principle holds true with
division. π₯ divided by π¦ to the π power is
equal to π₯ to the π power over π¦ to the π power. π₯ over π¦ cubed equals π₯ cubed
over π¦ cubed. And then, π₯ to the zero power is
one and π₯ to the first power is π₯. Using these rules, weβre ready to
jump in and simplify some expressions.

Simplify π₯ to the 12th power times
π₯ to the eighth power over π₯ times π₯ to the 11th power, where π₯ cannot be equal
to zero.

This is our expression. Looking at this expression, we
should recognize that when we have the same base π₯ to the π power times π₯ to the
π power will be equal to π₯ to the π plus π power. In our numerator, that means we can
add the exponents 12 and eight to get 20. In our denominator, we see an
exponent of 11. And we also know that π₯ is equal
to π₯ to the first power. And that means we can rewrite π₯ to
the first power times π₯ to the 11th power as π₯ to the 12th power.

And from there we think, will π₯ to
the π power over π₯ to the π power is equal to π₯ to the π minus π power. And that means we can take the
power in the numerator, 20, and subtract the power in the denominator, 12. When we simplify that, we get π₯ to
the eighth power. Since we do not know what π₯ is, we
canβt calculate the final answer, but π₯ to the eighth power is in its simplest
form.

Before we move on, I want to point
out one other way you couldβve solved this problem. In the numerator, we have π₯ to the
12th power. And in the denominator, we have π₯
to the first power times π₯ to the 11th power. If you recognize that one plus 11
equals 12, then you wouldβve seen that the numerator and the denominator both had a
factor of π₯ to the 12th power which cancel each other out. Either way, you end up with the
simplest form π₯ to the eighth power.

In our next example, weβre given an
equation, and weβre going to use that equation to help us simplify and then
calculate another expression.

Given five to the π₯ power equals
two, determine the value of 25 to the π₯ power.

We have the equation five to the π₯
power equals two and the expression 25 to the π₯ power. We want to know what the value of
that expression is. At first glance, we notice that
five and 25 are not the same base. And that makes it difficult to do
any kind of simplification. But when we think about that 25 in
relation to five, we could say that 25 equals five squared. In our expression, in place of 25,
we can write five squared so that we have five squared to the π₯ power. Which reminds us that π₯ to the π
power to the π power equals π₯ to the π times π power, which would be five to the
two times π₯ power.

But at this point, we still may be
puzzled because we only know what five to the π₯ power is; we donβt know what five
to the two π₯ power is. However, we can do some
regrouping. We know that five to the two π₯
power will also be equal to five to the π₯ power squared. What weβre saying is, five squared
to the π₯ power will have the same value as five to the π₯ power squared. This is true because theyβre both
equal to five to the two π₯ power.

At this point, we know what five to
the π₯ power is equal to. Since five to the π₯ power equals
two, we can calculate five to the π₯ power squared as two squared, which equals
four. Given that five to the π₯ power
equals two, the value of 25 to the π₯ power is four.

In our next example, weβll need to
simplify an expression to help us calculate its numerical value.

Given that π§ equals negative
one-fourth, find the numerical value of negative eight π§ to the fourth power times
π§ squared all over 16π§ cubed.

Starting with this expression, we
want to get it in its simplest form before we try and calculate its numerical
value. In simplifying, we want to get rid
of these parentheses. We remember that π₯ times π¦ to the
π power equals π₯ to the π power π¦ to the π power so that we have negative eight
to the fourth power times π§ to the fourth power times π§ squared. When we bring over the denominator
16π§ cubed, we can do some canceling between the numerator and the denominator. π₯ to the π power over π₯ two to
the π power equals π₯ to the π minus π power.

Since we have π§ to the fourth
power in the numerator and π§ cubed in the denominator, the π§ cubed in the
denominator cancels out and the π§ to the fourth power in the numerator becomes π§
to the first power. Four minus three is one. There are no more π§-variables in
the denominator, but in the numerator we have π§ to the first power times π§
squared. We can simplify that by writing it
as π§ cubed; one plus two is three. Now, we have negative eight to the
fourth power times π§ cubed over 16.

We can no longer simplify that
π§-variable. But it is worth considering if we
can simplify the negative eight to the fourth power over 16. I know that negative eight could
have a factor of negative four and two. If I write negative eight in this
way, we would then have negative four to the fourth power times two to the fourth
power times π§ cubed over 16, which is helpful to ask because two to the fourth
power is 16. So, the factor of 16 in the
numerator and the factor of 16 in the denominator cancel out.

We now have negative four to the
fourth power times π§ cubed. And weβre ready to plug in negative
one-fourth for π§. Negative four to the fourth power
times negative one-fourth to the third power. If we have π₯ over π¦ to the π
power, thatβs equal to π₯ to the π power over π¦ to the π power. If we break up this fraction, weβll
have negative four to the fourth power times negative one cubed over four cubed. I wanna break up this negative four
to the fourth power one final time so that we have negative one to the fourth power
times four to the fourth power.

Negative one to the fourth power
equals one. And since we have four cubed in the
denominator and four to the fourth power in the numerator, we can say that weβll
have four to the first power in the numerator. We subtract those exponents. Four minus three is one. Four to the first power equals
four. And the final operation we need to
do is negative one cubed. Thatβs negative one times negative
one, which is positive one, times negative one, which will be negative. So, our final answer here, the
numerical value for this expression is negative four.

Again, in the next example, weβll
have to do a significant amount of simplifying before we can answer the
question.

Which of these equations is
satisfied by π₯ equals two-thirds cubed squared and π¦ equals three times
three-halves to the fifth power minus three-halves to the sixth power? (A) π₯ equals π¦, (B) π₯ equals π¦
squared, (C) π₯ equals one over π¦, or (D) π₯ equals one over π¦ squared.

Letβs start with π₯. π₯ equals two-thirds cubed
squared. Using the power to a power rule, we
could rewrite this as two-thirds to the sixth power. And from there using another power
rule, we can simplify further to say two to the sixth power, three to the sixth
power. Now that this value is in its
simplest form, we donβt need to calculate yet. We can just leave it as two to the
sixth power over three to the sixth power.

π¦ equals three times three-halves
to the fifth power minus three-halves to the sixth power. And at first, itβs not immediately
clear what can be done here. But one thing we can do is break up
this three-halves to the sixth power, using the principle that π₯ to the π power
times π₯ to the π power equals π₯ to the π plus π power. We know that six equals five plus
one. And that means three-halves to the
first power times three-halves to the fifth power is equal to three-halves to the
sixth power. And then we have three times
three-halves to the fifth power minus three-halves to the first power times
three-halves to the fifth power.

And looking at these fractions,
again, it might not seem clear what we should do. But imagine, if we replace this
three-halves with the variable π₯. It becomes a little bit easier to
see that both of these terms have a factor of three-halves to the fifth power. If you undistributed π₯ to the
fifth power, youβd have π₯ to the fifth power times three minus π₯ to the first
power. We want to follow this principle
using three-halves to the fifth power, which means weβll have three-halves to the
fifth power times three minus three-halves to the first power.

Now, three-halves to the first
power is just three-halves. We wanna do the subtraction within
the brackets. And that means weβll need to
rewrite three with the denominator of two. Three is equal to six over two. Six-halves minus three-halves
equals three-halves. And now, we have three over two to
the fifth power times three over two, which is three over two to the first
power. And weβre able to use this same
principle to make this three-halves to the sixth power, which weβll rewrite as three
to the sixth power over two to the sixth power.

π¦ equals three to the sixth power
over two to the sixth power and π₯ equals two to the sixth power over three to the
sixth power. What weβre seeing is that π₯ and π¦
are the inverse of one another. When weβre working with fractions,
the inverse switches the numerator and the denominator. And if π₯ and π¦ are the inverse of
one another, then π₯ equals one over π¦.

If you wanted to check that this
was true, we have a statement that says two to the sixth power over three to the
sixth power is equal to one over three to the sixth power over two to the sixth
power. Thatβs saying one divided by three
to the sixth power over two to the sixth power. We could rewrite that as one
divided by three to the sixth power over two to the sixth power. And we know dividing by a fraction
is the same as multiplying its reciprocal. So that would be one times two to
the sixth power over three to the sixth power, which is true. For these equations, π₯ is equal to
one over π¦.

In our final example, weβll be
working backwards from something thatβs already simplified to find its previous
form.

Complete the following: 14π₯ cubed
π¦ to the fifth power over something is equal to two π₯ over six π¦.

When we look at this equation,
whatβs on the right is the simplified form of the fraction on the left. Weβll need to work backwards to
find out what the denominator of the first fraction was. Since weβre working backwards, Iβm
gonna flip the order so that we have two π₯ over six π¦ is equal to 14π₯ cubed π¦ to
the fifth power. Even though we have variables and
their coefficients, the principles of keeping fractions proportional still
apply. And that means in the numerator, if
we went from two to 14, we would have to multiply by seven. But if we multiplied by seven in
our numerator, weβd have to multiply by seven in our denominator to keep the values
equal.

Six times seven is 42. Going back to our numerator, we
have an π₯ going to an π₯ cubed. Here, that means weβve multiplied
by π₯ squared. And if we multiply by π₯ squared in
the numerator, we have to multiply by π₯ squared in the denominator. Since the denominator doesnβt have
an π₯ squared, itβs one times π₯ squared, which is π₯ squared. And then in our numerator, we also
have π¦ to the fifth. We donβt have that on the left. So thatβs just going to be
multiplying by π¦ to the fifth. We do have a π¦ in the
denominator. So, weβre multiplying the
denominator by π¦ to the fifth, which is π¦ times π¦ to the fifth, which is π¦ to
the sixth power.

Because we donβt generally work in
that direction, itβs probably a good idea to check the result. Both 14 and 42 are divisible by
seven. That gives two-sixths. The π₯ squared in the denominator
is canceled out and leaves us with π₯ to the first power in the numerator. When we look at our π¦-variables,
the π¦ to the fifth power in the numerator cancels out. And weβre left with π¦ to the first
power in the denominator, which is two π₯ over six π¦ and confirms 42π₯ squared π¦
to the sixth power as the missing denominator.

When it comes to the key points
here, simplifying expressions allows us to write an expression in the most compact
and efficient manner without changing the value of the expression. And when those expressions have
exponents, we can use these basic rules to help us simplify.