Lesson Video: Power and Exponents over the Real Numbers | Nagwa Lesson Video: Power and Exponents over the Real Numbers | Nagwa

# Lesson Video: Power and Exponents over the Real Numbers Mathematics • Second Year of Preparatory School

## Join Nagwa Classes

In this video, we will learn how to evaluate real numbers raised to positive and negative integer and zero powers and solve simple exponential equations.

17:32

### 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. Letβs 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 πth 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. Letβs use this result in an example.

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

Letβs 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 thatβs eight multiplied by eight, multiplied by 10, because thatβs 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 letβs 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 Iβve 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? Weβd 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 letβs 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. Thatβs going to be one over two. So two to the negative one power is one over two.

Letβs repeat this one more time. We can do this again by dividing the subsequent term, thatβs 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.

Letβs see this in an example.

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

Letβs begin by substituting π₯ equals root three into π₯ to the zero power. At this point, letβs 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 letβs 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, letβs recall the law for negative exponents. That is, π to the negative π power equals one over π to the πth 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. Thatβs option (B).

Letβs 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.

Letβs 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 πth power equals one over π to the πth 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. Letβs 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 weβve seen how to evaluate real numbers raised to zero or negative exponents, letβs 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.

Letβs 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. Letβs now calculate root two over three root two squared. Thatβs 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. Thatβs 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, letβs 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 πth power is one over π to the πth 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.

Letβs 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 weβve 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 thatβs just three. So we have one over nine multiplied by three. Thatβs three over nine, which is one-third.

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

Letβs 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.

Letβs 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. Letβs 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 letβs 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.

Letβs see if we can simplify this right-hand side of the equation. To help us do this, letβs 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 ππ 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 weβve seen how to evaluate real numbers raised to positive, zero, and negative exponents and how to solve simple exponential equations.

Letβs 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 πth 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.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions