### Video Transcript

In this video, we will practice
evaluating algebraic expressions with variable exponents. We might first want to remind
ourselves what an expression is. Expressions are numbers, symbols,
and operators grouped together to show the value of something. For example, if we have the
equation three plus five to the π₯ power equals 128, three plus five to the π₯ power
is an expression. In this expression, we have five to
the π₯ power. We have an unknown value as the
power. In this video, we will specifically
be looking at algebraic expressions where solving them will include solving for a
variable as the power or the index, exponents where the power is unknown.

But before we get to that, we need
to do a quick review of the exponent rules and working with exponents. When weβre multiplying exponents,
π₯ to the π power times π₯ to the π power equals π₯ to the π plus π power. When weβre dividing exponents, π₯
to the π power divided by π₯ to the π power equals π₯ to the π minus π
power. Taking a power of a power, π₯ to
the π power to the π power equals π₯ to the π times π power. And then we have the power of a
product, which would be something like π₯ times π¦ to the π power, which would
yield π₯ to the π power times π¦ to the π power.

And then we have the power of a
fraction. If we have π₯ divided by π¦ to the
π power, weβll have π₯ to the π power divided by π¦ to the π power. Next, we have a zero exponent. π₯ to the zero power equals
one. Anything taken to the zero power
equals one. And for our negative exponents, π₯
to the negative π power is equal to one over π₯ to the π power. Finally, if we have a fractional
exponent, sometimes called rational powers, we would have π₯ to the one over π
power. And that is equal to the πth root
of π₯. Using these rules, weβll be able to
simplify and solve for variable exponents. Letβs consider some examples.

Evaluate three plus five to the π₯
power if π₯ equals two.

In this case, we have the
expression three plus five to the π₯ power. In order to evaluate this
expression, we substitute the number two in for π₯ so that we have three plus five
squared. Five squared is five times five,
which is 25. We add three to 25 to get 28, which
tells us that three plus five to the π₯ power is equal to 28 when π₯ equals two.

In our next example, weβll consider
a case where weβre dealing with two different variables.

Evaluate π₯ to the π§ power minus
π§ to the π₯ power, where π₯ equals four and π§ equals three.

Weβre given the algebraic
expression π₯ to the π§ power minus π§ to the π₯ power. Itβs possible at this point that
you might think π₯ to the π§ power minus π§ to the π₯ power must be zero. However, this is not true. π₯ to the π§ power does not equal
π§ to the π₯ power. They are not additive inverses of
each other. And we can confirm this by plugging
in the values that we were given for π₯ and π§. Since we know that π₯ equals four
and π§ equals three, our first term will be four cubed and our second term will be
three to the fourth power. Four cubed is equal to 64, and
three to the fourth power equals 81. This expression would then be 64
minus 81, which equals negative 17. And we see here that four cubed, π₯
to the π§ power, does not have the same value as three to the fourth power, π§ to
the π₯ power.

In our third example, weβll look at
two fractional values being multiplied together when they have different variable
exponents.

Calculate the numerical value of
five-fourths to the π¦ power times five-fourths to the π₯ power when π₯ is equal to
negative seven and π¦ equals four.

We have the expression five-fourths
to the π¦ power times five-fourths to the π₯ power. When we look at this expression, we
should notice that both of our terms have five over four as their base. And theyβre being multiplied
together. This should make us think of our
exponent rule π₯ to the π power times π₯ to the π power is equal to π₯ to the π
plus π power. This means we can rewrite our
expression with the base five over four and a new exponents of π¦ plus π₯. This step simplifies our expression
into one term. And from there, weβll substitute
negative seven in for π₯ and four in for π¦. We would then have five-fourths to
the four plus negative seven power, which we can simplify to the four minus seven
power.

But four minus seven is negative
three, which leaves us with five-fourths to the negative three power. We can remember the power of a
fraction rule, which tells us that π₯ over π¦ to the π power is equal to π₯ to the
π power over π¦ to the π power. When we distribute this negative
three power, we get five to the negative three power over four to the negative three
power. And we also know that π₯ to the
negative π power is equal to one over π₯ to the π power. A corollary of this tells us one
over π₯ to the negative π power is equal to π₯ to the π power, which means that if
we have a negative exponent in the denominator, we can move it to the numerator and
make it positive. Five to the negative three power
over four to the negative three power can be rewritten as four cubed over five
cubed, which equals 64 over 125.

In our fourth example, we wonβt be
told exactly what our exponent values are. Instead, weβll have to use given
information to calculate the values of variables in the exponent.

Given that two to the π₯ power
equals eight to the π¦ power which equals 512, determine the value of π₯ minus
π¦.

Before we can solve for π₯ minus
π¦, weβll have to determine what π₯ and π¦ are equal to. This statement tells us that two to
the π₯ power equals 512 and eight to the π¦ power also equals 512. We could break this up into two
separate statements, like this. And thinking about solving each of
these, if 512 is equal to two to some power and eight to some power, we know that
512 will be divisible by both two and eight. Since eight is the larger of the
two factors, one thing I would do is to try and divide 512 by eight. When I do that, I see that eight
times 64 equals 512. And I already know that 64 equals
eight times eight.

This shows me that 512 is equal to
eight cubed. In our statement, where we see 512,
we can substitute eight cubed since 512 equals eight cubed. And then we get a new
statement. It says eight to the π¦ power
equals eight cubed. And based on this information, we
can say that π¦ must be equal to three. Now, weβve already found out that
512 equals eight cubed. But when we have the statement two
π₯ equals eight cubed, weβre still not in a position to solve for π₯. But we could be thinking that eight
is divisible by two. We know that two times four equals
eight. And then two times two equals
four. And that means we could rewrite the
value eight as two cubed, which means 512 equals two cubed cubed.

Taking what we know about power to
a power, π₯ to the π power to the π power equals π₯ to the π times π power. Three times three is nine, which
means 512 is equal to two to the ninth power. And our new statement says two to
the π₯ power equals two to the ninth power. Therefore, π₯ must be equal to
nine. Weβre now ready to solve the
expression π₯ minus π¦. We know π₯ equals nine and π¦
equals three. So we can say that π₯ minus π¦
equals six; nine minus three equals six.

In our final example, again, weβll
have to solve for the value of a missing exponent. And then weβll use that information
to determine the value of an expression.

Given that two to the π₯ power
equals four, determine the value of eight to the π₯ minus one power times two to the
two π₯ plus six power times one-half to the two π₯ power.

Weβve been given a statement and
then an expression to solve for. In order to solve the expression,
weβll need to take the statement two to the π₯ power equals four. Using this information, we can
solve for π₯. We want to rewrite four with a base
of two. Since we know that four equals two
squared, we can substitute two squared into this statement in place of four, which
allows us to see that two to the π₯ power equals two squared. Therefore, π₯ must be equal to
two. Once we know that, weβre ready to
consider our expression, eight to the π₯ minus one power times two to the two π₯
plus six power times one-half to the two π₯ power.

In this expression, everywhere we
see the variable π₯, weβll substitute the value two. Eight to the two minus one power
would be eight to the first power, which is eight. Two to the two times two plus six
power β β two times two is four plus six is 10. Weβll have two to the tenth
power. And one-half to the two times two
power is one-half to the fourth power. Now, at this point, it is
completely possible to plug this into the calculator or try and solve for all of
these values.

However, there is a way we can
simplify a bit first if we recognize that all three of these factors can be
rewritten with a base of two. Just like we rewrote four as two
squared, we could rewrite eight as two cubed. Two times two times two equals
eight. Two to the tenth power already has
a base of two. And now, you might be wondering,
βbut how will we write one-half as a base of two?β Remember that π₯ to the negative π
power equals one over π₯ to the π power. This means we could think of one
over two as one over two to the first power. And we can rewrite this as two to
the negative first power.

Instead of one-half to the fourth
power, weβll have two to the negative first power to the fourth power. And because we know a power of a
power rule, π₯ to the π power to the π power equals π₯ to the π times π power,
two to the negative one power to the fourth power can be rewritten as two to the
negative four power. And one final step for
simplification, π₯ to the π power times π₯ to the π power equals π₯ to the π plus
π power. Since all of our factors have a
base of two and theyβre being multiplied together, we take the base of two and then
we add all the exponents together. It will be two to the three plus 10
plus negative four power. Three plus 10 is 13, minus four is
nine, two to the ninth power. And two to the ninth power equals
512.

Before we finish, letβs remember a
few key points for evaluating expressions with variable exponents. When evaluating algebraic
expressions with variable exponents, use the rules of exponents to simplify as much
as possible. In addition to that, where
possible, rewrite values so that they have the same base. Then use power of a power rule to
further evaluate.