### Video Transcript

In this lesson, we will learn how
to evaluate real numbers raised to positive and negative integer and zero powers and
solve simple exponential equations. Letβs begin by recalling what it
means to evaluate a power.

For a power with base π, where π
is in the set of real numbers not including zero, and exponent π, where π is in
the set of positive integers, then π raised to the πth power is π multiplied by
itself π times. With this definition, we can
calculate not only powers where the base is an integer or a rational number, but
also where the base is a real number. So this includes irrational numbers
too, such as square roots.

If we take the square root of π,
where π is a positive integer, squaring this just gives us π, since squaring and
square rooting are inverse operations. Another thing to notice is that if
we take π squared and square root it, we get π or negative π. However, we can say that the square
root of π squared is π if we know that π is greater than zero.

So one result to remember is we can
simplify expressions with squares and square roots in an expression with base π,
where π is greater than zero, using the following rules. The square root of π squared
equals root π multiplied by root π, which equals π. And the square root of π
multiplied by π equals the square root of π squared, which equals π. Letβs use this result in an
example.

Determine the value of π₯
squared given that π₯ equals eight root 10.

Letβs begin this question by
substituting π₯ equals eight root 10 into π₯ squared. So we have that π₯ squared
equals eight root 10 squared. We know that from the
definition of a power that this is the same as eight root 10 multiplied by eight
root 10. Now, remembering that eight
root 10 just means eight multiplied by root 10, we can write this as eight
multiplied by eight multiplied by root 10 multiplied by root 10. Then, using the general result
that root π multiplied by root π equals π, we get 64, because thatβs eight
multiplied by eight, multiplied by 10, because thatβs root 10 multiplied by root
10. And of course that gives us
640.

So that was an example of raising a
real number to a positive integer power. So now letβs also look at raising a
real number to a zero power.

Consider powers of two: two to the
first power, two to the second power, two to the third power, two to the fourth
power, and so on. These are, respectively, two, four,
eight, and 16. You can see how Iβve written this
in a sequence increasing each power by one. And therefore each term is
multiplied by two to get to the next term. But what if we go backward in the
sequence? Weβd have two to the zero power,
which we can find by dividing the two to the first power term by two. That gives us one, so two to the
zero power is one.

Now letβs consider moving backward
again. That will give us two to the
negative one power, which we can find by dividing the subsequent term, two to the
zero power, which is one, by two. Thatβs going to be one over
two. So two to the negative one power is
one over two.

Letβs repeat this one more
time. We can do this again by dividing
the subsequent term, thatβs two to the negative one power, which is one-half, by
two. And we find that two to the
negative two power must be one over four. One thing to notice here is that we
can actually rewrite the value of two to the negative one power, which is one over
two, as one over two to the first power. And we can rewrite two to the
negative two power as one over two to the second power.

In fact, we can generalize this in
the following result. The law for zero exponents is π to
the zero power equals one, where π is in the set of real numbers not including
zero. The law for negative exponents is
π to the negative π power equals one over π to the π power, where π is in the
set of real numbers without zero and π is an integer.

Letβs see this in an example.

If π₯ equals root three, find π₯ to
the zero power.

Letβs begin by substituting π₯
equals root three into π₯ to the zero power. At this point, letβs recall the law
for zero exponents. That is, π to the zero power
equals one, where π is in the set of real numbers without zero. Therefore, root three to the zero
power is just one.

Now letβs see an example where we
raise a real number to a negative power.

If π₯ is equal to root four over
root two, which of the following is equal to π₯ to the negative one power? (A) Negative root two over two, (B)
root two over two, (C) negative root two, (D) one, or (E) root two.

We can start by substituting π₯
equals root four over root two into π₯ to the negative one power.

Now, letβs recall the law for
negative exponents. That is, π to the negative π
power equals one over π to the πth power, where π is in the set of all real
numbers without zero and π is an integer. So, using this rule, root four over
root two raised to the negative one power is one over root four over root two.

We know that when we have a
fraction on the denominator like this, we can just write this as root two over root
four. Now, we can actually simplify this
answer because we can calculate root four to be two. So our answer is root two over
two. Thatβs option (B).

Letβs have a look at another
question where we raise a real number, a square root to be specific, to a negative
integer power.

If π₯ equals root two, find π₯
to the negative fourth power.

Letβs start by substituting π₯
equals root two into π₯ to the negative fourth power. Now, to calculate root two to
the negative fourth power, we can apply the law for negative exponents. That is, π to the negative
πth power equals one over π to the πth power, where π is in the set of real
numbers without zero and π is an integer. So this tells us that we can
write root two to the negative fourth power as one over root two to the fourth
power.

Now, root two to the fourth
power is something we can calculate. Letβs expand this power. We know that evaluating a power
means that we multiply the base by itself π times. So root two to the fourth power
is the same as root two multiplied by root two multiplied by root two multiplied
by root two.

At this point, we can recall
the rule that root π multiplied by root π is just π for π greater than
zero. Therefore, we can simplify this
to be two multiplied by two, and that just gives us four. So our denominator is four, and
our answer is therefore one over four.

Now weβve seen how to evaluate real
numbers raised to zero or negative exponents, letβs look at how we can evaluate real
numbers raised to both positive and negative integer exponents in the same
expression.

If π₯ equals root two over
three, π¦ equals one over root two, and π§ equals root three over three, find
π₯π¦ squared π§ to the negative two power in its simplest form.

Well, to find the value of π₯π¦
squared multiplied by π§ to the negative two power, we need to begin by
substituting the values for π₯, π¦, and π§ into the expression. Doing so gives us root two over
three multiplied by one over root two squared multiplied by root three over
three to the negative two power.

Letβs simplify what we have
inside this parentheses by multiplying root two over three by one over root
two. That gives us root two over
three root two. Letβs now calculate root two
over three root two squared. Thatβs going to be root two
over three root two multiplied by root two over three root two. Then, multiplying the
numerators together, we get two. And multiplying the
denominators together, we get nine multiplied by two. Thatβs because we can firstly
multiply the threes to give nine and then the root twos to give us two. Because we know that root two
multiplied by root two just gives us two.

Now, nine multiplied by two
gives us 18. And then we can divide both the
numerator and denominator by two to simplify this. That gives us one over
nine. Another approach we could use
to evaluate this would be to cancel the root two on the numerator with the root
two on the denominator and evaluate one-third squared.

Now, letβs consider root three
over three to the negative two power. We firstly recall the law for
negative exponents, which states that π to the negative πth power is one over
π to the πth power, for π in the set of real numbers without zero and π an
integer. This means that we can write
root three over three to the negative two power as one over root three over
three to the second power.

Letβs evaluate the denominator,
root three over three to the second power. This is root three over three
multiplied by root three over three. Then, multiplying the
numerators, root three by root three, gives us three. And multiplying the
denominators, three multiplied by three, gives us nine. And three over nine simplifies
to one over three. So now weβve calculated π₯π¦
squared π§ to the negative two power to be one over nine multiplied by one over
one over three.

Now, we can actually rewrite
one over one-third, because this is simply one divided by a third, and thatβs
just three. So we have one over nine
multiplied by three. Thatβs three over nine, which
is one-third.

Weβre going to see some examples of
solving exponential equations with unknown exponents and bases. So now we consider how to find
unknown exponents.

Letβs say we have that three to the
fourth power is equal to three raised to the π¦ power. Then, seen as the bases are the
same, we can conclude that the powers must be the same too, so that π¦ must be equal
to four. If we compare two powers with the
same exponent and an unknown base, we may conclude that the bases are the same
too. However, this only works when the
powers are equal and odd, because with an even exponent we could have that one of
the bases is the negative of the other. For instance, negative two to the
fourth power is 16. Likewise, two to the fourth power
is also 16. So this is why even exponents do
not follow the same rule.

If the exponents are equal, we can
say that the moduli of the bases are equal. For example, if π to the fourth
power is equal to two to the fourth power, then the modulus of π equals two so that
π equals two or negative two.

Letβs summarize these results. Equating powers with the same base
or the same exponent. If π to the π power equals π to
the π power, where π is in the set of real numbers without negative one, zero, or
one, then π is equal to π. If π to the π power is equal to
π to the π power, then π equals π when π is in the set of numbers one, three,
five, and so on. And the modulus of π is equal to
the modulus of π when π is in the set of numbers two, four, six, and so on. Letβs apply this rule in an
example.

Find the value of π₯ in the
equation two to the π₯ minus three power equals root two to the π₯ plus one
power.

To find the value of π₯ in this
equation, we can use the rule π to the π power equals π to the π power,
where π is in the set of real numbers without negative one, zero, or one, then
π is equal to π. However, this rule only works
when both sides of the equation have the same base. The left side of our equation
has base two and the right side base root two. So letβs aim to rewrite the
right side with a base of two also.

One rule that we can recall
here is that we can write the square root of π as π to the half power. So root two can be written as
two to the half power. So our original equation can be
written as two to the π₯ minus three power equals two to the half power to the
π₯ plus one power.

Letβs see if we can simplify
this right-hand side of the equation. To help us do this, letβs
recall how we can raise a number to a power to another power. We use the rule π₯ to the π
power raised to the π power is π₯ to the ππ power. Therefore, two to the half
power raised to the π₯ plus one power is two to the half multiplied by π₯ plus
one power. Since the bases are now the
same, we can equate the exponents using the rule that we wrote down here. π₯ minus three equals half
multiplied by π₯ plus one.

We can begin to solve this by
multiplying both sides of the equation by two, then adding six to both sides,
and finally subtracting π₯ from both sides, to give us π₯ equals seven. Therefore, the value of π₯ for
this equation is seven.

So now weβve seen how to evaluate
real numbers raised to positive, zero, and negative exponents and how to solve
simple exponential equations.

Letβs now summarize the main points
from this lesson.

We can use the fact that root π
squared is root π multiplied by root π, and that gives us π, to simplify radicals
raised to positive integers. The law for zero exponents states
that π to the zero power equals one, where π is in the set of real numbers
excluding zero. The law for negative exponents
states that π to the negative π power equals one over π to the πth power, where
π is in the set of real numbers excluding zero and π is in the set of integers
excluding zero. And we can use this law to simplify
real numbers raised to negative powers.

We can use rules to find unknowns
in simple exponential equations. These are if π to the π power
equals π to the π power, where π is in the set of real numbers without negative
one, zero, or one, then π is equal to π. If π to the π power equals π to
the π power, then π is equal to π when π is in the set of odd integers. And the modulus of π is equal to
the modulus of π when π is in the set of positive integers.