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Lesson Explainer: Power and Exponents over the Real Numbers Mathematics • 9th Grade

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

Recall that to evaluate a power, we multiply the base the number of times indicated by the exponent. Let’s recap the definition below.

Definition: Evaluating a Power

For a power with base π‘Ž, π‘Žβˆˆβ„βˆ’0, and exponent 𝑛, π‘›βˆˆβ„€οŠ°, then π‘Ž=π‘ŽΓ—π‘ŽΓ—π‘ŽΓ—β€¦Γ—π‘Žο‡Œο†²ο†²ο†²ο†²ο‡ο†²ο†²ο†²ο†²ο‡Ž.times

Using this definition, we can evaluate not only powers where the base is an integer or a rational number, but also where the base is a real number.

One such type of real number we might evaluate is a square root. Recall that a square root is a number written in the form βˆšπ‘Ž, π‘Žβˆˆβ„€οŠ°. We can use the inverse operation, squaring, to derive two key results.

The first is that if you square βˆšπ‘Ž, then you get π‘Ž, since squaring and square rooting are inverse operations (i.e., ο€Ίβˆšπ‘Žο†=π‘ŽοŠ¨, π‘Ž>0).

The second is that if you square root π‘ŽοŠ¨, you either get π‘Ž or βˆ’π‘Ž, depending on whether π‘Ž is negative or positive. This is because when we square a negative number, say βˆ’3, we get a positive number, 9. However, if we square the additive inverse of βˆ’3, which is 3, we also get 9. This means when we square root a number squared, the result is positive, but the original number might have been negative. To avoid confusion, we then say that βˆšπ‘Ž=|π‘Ž|. However, if we know π‘Ž>0, then we can say βˆšπ‘Ž=π‘ŽοŠ¨.

Expanding on these definitions and using the definition of a power, we can say that ο€Ίβˆšπ‘Žο†=βˆšπ‘ŽΓ—βˆšπ‘ŽοŠ¨ and βˆšπ‘Ž=βˆšπ‘ŽΓ—π‘ŽοŠ¨. Therefore, we get the following definition.

Definition: Simplifying Expressions with Squares and Square Roots

For an expression with a base π‘Ž, where π‘Ž>0,

  • ο€Ίβˆšπ‘Žο†=βˆšπ‘ŽΓ—βˆšπ‘Ž=π‘ŽοŠ¨,
  • βˆšπ‘ŽΓ—π‘Ž=βˆšπ‘Ž=π‘ŽοŠ¨.

In our first example, we will apply what we know about evaluating powers and calculating squares of radicals to determine the value of an unknown.

Example 1: Raising a Real Number to a Positive Integer Power

Determine the value of π‘₯ given that π‘₯=8√10.

Answer

To find the value of π‘₯, we need to substitute π‘₯=8√10 into π‘₯. Doing so gives us π‘₯=ο€»8√10.

We know from the definition of a power that ο€»8√10ο‡οŠ¨ is the same as saying 8√10Γ—8√10. Therefore, we get π‘₯=8√10Γ—8√10.

To calculate this, we need to use βˆšπ‘ŽΓ—βˆšπ‘Ž=π‘Ž, giving us π‘₯=8Γ—8Γ—βˆš10Γ—βˆš10=64Γ—10=640.

Therefore, the value of π‘₯ is 640.

As well as raising real numbers to the power of a positive integer, we can also raise them to the power of zero or to the power of a negative integer.

If we consider the pattern of powers of 2, we can see that as we increase the power by 1, we multiply by 2, as follows: 2,2,2,2,…2,4,8,16,β€¦οŠ§οŠ¨οŠ©οŠͺ

If we move backward in the sequence, we can therefore deduce that 2 must be 2 divided by 2. This then gives us 2÷2=1. So, 2=1.

Similarly, if we move backward again, we can deduce the 2 must be 2 divided by 2. This then gives us 1÷2=12. So, 2=12.

Again, if we continue moving backward, we can deduce that 2=14 and 2=18 and so on. If we write out the sequence again, we get the following: 2,2,2,2,2,2,2,2,…18,14,12,1,2,4,8,16,β€¦οŠ±οŠ©οŠ±οŠ¨οŠ±οŠ§οŠ¦οŠ§οŠ¨οŠ©οŠͺ

Notice, that if we rewrite the denominators in the first three terms as powers of 2, we get the following: 2,2,2,2,2,2,2,2,…12,12,12,1,2,4,8,16,β€¦οŠ±οŠ©οŠ±οŠ¨οŠ±οŠ§οŠ¦οŠ§οŠ¨οŠ©οŠͺ

We can then say that 2=12, 2=12, and 2=12. Continuing the sequence in the negative direction, we can generalize that 2=12, π‘›βˆˆβ„€.

These results do not only hold for a base of 2 but also for any real number except zero. We can generalize these in the laws below.

Laws: Rules for Zero and Negative Exponents

  • The law for zero exponents is π‘Ž=1, where π‘Žβˆˆβ„βˆ’0.
  • The law for negative exponents is π‘Ž=1π‘Ž, where π‘Žβˆˆβ„βˆ’0 and π‘›βˆˆβ„€.

In the following example, we will use the law for zero exponents to determine the value of an expression involving powers of radicals.

Example 2: Raising a Real Number to a Zero Power

If π‘₯=√2, find π‘₯.

Answer

To determine the value of π‘₯, we can substitute π‘₯=√2, giving us π‘₯=ο€»βˆš2.

We know that π‘Ž=1, where π‘Žβˆˆβ„βˆ’0; therefore, π‘₯=1.

Therefore, the value of π‘₯ is 1.

In the next example, we will use the law for negative exponents to determine the value of an expression involving powers of radicals.

Example 3: Raising a Real Number to a Negative Integer Power

If π‘₯=√2, find π‘₯οŠͺ.

Answer

To find the value of π‘₯οŠͺ when π‘₯=√2, we first need to substitute. This gives us π‘₯=ο€»βˆš2.οŠͺοŠͺ

Next, to simplify ο€»βˆš2ο‡οŠ±οŠͺ, we use the law for negative exponents, which states that π‘Ž=1π‘Ž, where π‘Žβˆˆβ„βˆ’0 and π‘›βˆˆβ„€. This then gives us ο€»βˆš2=1ο€»βˆš2.οŠͺοŠͺ

We can use the rule βˆšπ‘ŽΓ—βˆšπ‘Ž=π‘Ž, π‘Ž>0, to evaluate the denominator by expanding the power, giving us ο€»βˆš2=√2Γ—βˆš2Γ—βˆš2Γ—βˆš2=2Γ—2=4.οŠͺ

Substituting this into the denominator, we get 1ο€»βˆš2=14.οŠͺ

Therefore, the value of π‘₯οŠͺ when π‘₯=√2 is 14.

In the next example, we will discuss how to evaluate an expression with positive and negative exponents.

Example 4: Evaluating an Expression by Substituting Surds

If π‘₯=√32, 𝑦=1√2, and 𝑧=√23, then find the value of π‘₯+(𝑦+𝑧)βˆ’π‘§οŠ¨οŠ¨οŠͺ in its simplest form.

Answer

To find the value of π‘₯+(𝑦+𝑧)βˆ’π‘§οŠ¨οŠ¨οŠͺ, we start by substituting π‘₯=√32, 𝑦=1√2, and 𝑧=√23. Doing so gives us π‘₯+(𝑦+𝑧)βˆ’π‘§=ο€Ώβˆš32+ο€Ώ1√2+√23ο‹βˆ’ο€Ώβˆš23.οŠͺοŠͺ

We can simplify this further by expanding the brackets of the second term, giving us ο€Ώ1√2+√23=ο€Ώ1√2+2Γ—1√2Γ—βˆš23+ο€Ώβˆš23.

Canceling common factors of √2 in 2Γ—1√2Γ—βˆš23 gives us ο€Ώ1√2+√23=ο€Ώ1√2+2Γ—1Γ—13+ο€Ώβˆš23=ο€Ώ1√2+23+ο€Ώβˆš23.

Substituting this back into the expression, we get π‘₯+(𝑦+𝑧)βˆ’π‘§=ο€Ώβˆš32+ο€Ώ1√2+23+ο€Ώβˆš23ο‹βˆ’ο€Ώβˆš23.οŠͺοŠͺ

Next, we can use ο€Ίβˆšπ‘Žο†=π‘ŽοŠ¨ to evaluate the squared terms, giving us for the first, second, and fourth terms, respectively, ο€Ώβˆš32=√32Γ—βˆš32=ο€»βˆš32Γ—2=34,ο€Ώ1√2=1√2Γ—1√2=1Γ—1ο€»βˆš2=12,ο€Ώβˆš23=√23Γ—βˆš23=ο€»βˆš23Γ—3=29.

Substituting back into the expression, we get π‘₯+(𝑦+𝑧)βˆ’π‘§=34+12+23+29βˆ’ο€Ώβˆš23.οŠͺοŠͺ

Since ο€Ώβˆš23οŠͺ is √23Γ—βˆš23Γ—βˆš23Γ—βˆš23, then we can use βˆšπ‘ŽΓ—βˆšπ‘Ž=π‘Ž to simplify this, giving us √23Γ—βˆš23Γ—βˆš23Γ—βˆš23=29Γ—29=481.

Substituting back into the expression, we get π‘₯+(𝑦+𝑧)βˆ’π‘§=34+12+23+29βˆ’481.οŠͺ

To simplify further, we need to find a common factor for the denominators of the fractions. For the first four terms, a common factor is 36, giving us π‘₯+(𝑦+𝑧)βˆ’π‘§=2736+1836+2436+836βˆ’481=7736βˆ’481.οŠͺ

Finally, a common factor of 36 and 81 is 324, giving us π‘₯+(𝑦+𝑧)βˆ’π‘§=693324βˆ’16324=677324.οŠͺ

Therefore, the answer is 677324.

So far, we have considered how to simplify expressions with integer exponents. Next, we will discuss how to find an unknown exponent.

When comparing two powers with the same base, we can deduce that if the powers are equal, then their exponents must be equal as well. For example, if 3=3οŠͺ, then since the bases are the same, the exponents must be the same too, meaning 𝑦=4.

Similarly, if we compare two powers with the same exponent, then if they are equal and odd, the bases must be equal. For example, if π‘Ž=7, then since the exponents are the same, then the bases must be the same too, meaning π‘Ž=7. However, if we have an even exponent, then it is possible that the bases are not equal, since one could be the negative of the other. For example, (βˆ’2)=2, so if the exponents are equal, then the base could be βˆ’2 or 2 in order for them to be equal. This means that if the even exponents are equal, then we can say the moduli of the bases are equal. For example, if 𝑐=3οŠͺοŠͺ, then |𝑐|=3, so 𝑐=3 or 𝑐=βˆ’3. Let’s summarize these points in the rule below.

Rule: Equating Powers with the Same Base or the Same Exponent

  • If π‘Ž=π‘Žο‰οŠ, where π‘Žβˆˆβ„βˆ’[βˆ’1,0,1], then π‘š=𝑛.
  • If π‘Ž=π‘οŠοŠ, then π‘Ž=𝑏 when π‘›βˆˆ[1,3,5,…] and |π‘Ž|=|𝑏| when π‘›βˆˆ[2,4,6,…].

We can use these rules to solve problems where there are unknowns in the exponent. Let’s discuss how to do this in the following example.

Example 5: Solving a Simple Exponential Equation

Find the value of π‘₯ in the equation 3=181οŠ¨ο—οŠ±οŠ§.

Answer

To find the value of π‘₯ in the equation 3=181οŠ¨ο—οŠ±οŠ§, we can use the rule that states that if π‘Ž=π‘Žο‰οŠ, where π‘Žβˆˆβ„βˆ’[βˆ’1,0,1], then π‘š=𝑛.

This rule will only work if both sides of the equation have the same base. As such, since the base of the power on the left-hand side of the equation is 3, then we want to write the right-hand with a base of 3 also.

We know that 81 is a power of 3, since 81=3Γ—3Γ—3Γ—3. Therefore, 81=3οŠͺ. We can then say that the right-hand side is 181=13.οŠͺ

Next, we can use the law of negative exponents, which states π‘Ž=1π‘Ž, where π‘Žβˆˆβ„βˆ’0 and π‘›βˆˆβ„€βˆ’0, meaning that for 13οŠͺ, we get 13=3.οŠͺοŠͺ

Putting this into the right-hand side of the original equation, we then get 3=3.οŠ¨ο—οŠ±οŠ§οŠ±οŠͺ

Since the bases are now the same, we can equate the exponents, giving us 2π‘₯βˆ’1=βˆ’4.

Solving for π‘₯, we get 2π‘₯=βˆ’3π‘₯=βˆ’32.

Therefore, the value of π‘₯ is βˆ’32 for the equation 3=181οŠ¨ο—οŠ±οŠ§.

In this explainer, we have learned how to evaluate real numbers raised to positive, zero, and negative exponents and how to solve simple exponential equations. Let’s recap the key points.

Key Points

  • We can use the property βˆšπ‘ŽΓ—βˆšπ‘Ž=ο€Ίβˆšπ‘Žο†=π‘ŽοŠ¨ to simplify radicals raised to positive integers.
  • We can use the law for zero exponents to simplify real numbers raised to the power of zero. The law states that π‘Ž=1, where π‘Žβˆˆβ„βˆ’0.
  • We can use the law for negative exponents to simplify real numbers raised to negative powers. The law states that π‘Ž=1π‘Ž, where π‘Žβˆˆβ„βˆ’0 and π‘›βˆˆβ„€βˆ’0.
  • We can use the rules for when bases or exponents are equal for equal powers to find unknowns in simple exponential equations. The rules state that
    • if π‘Ž=π‘Žο‰οŠ, where π‘Žβˆˆβ„βˆ’[βˆ’1,0,1], then π‘š=𝑛,
    • if π‘Ž=π‘οŠοŠ, then π‘Ž=𝑏 when π‘›βˆˆ[1,3,5,…] and |π‘Ž|=|𝑏| when π‘›βˆˆ[2,4,6,…].

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