Lesson Explainer: Evaluating Expressions: Exponents | Nagwa Lesson Explainer: Evaluating Expressions: Exponents | Nagwa

Lesson Explainer: Evaluating Expressions: Exponents Mathematics

In this explainer, we will learn how to evaluate algebraic expressions with variable exponents or bases.

Recall that in algebraic expressions of the form 𝑎, 𝑎 is a real number called the base and 𝑛 is an integer called the exponent (or index, or power). We refer to these as exponential expressions. When describing 𝑎 in words, we say “𝑎 to the power of 𝑛,” or simply “𝑎 to the power 𝑛,” which is the mathematical way to describe 𝑛 copies of the number 𝑎 multiplied together. As an example, “𝑎 to the power of 4” means 𝑎=𝑎×𝑎×𝑎×𝑎.

Sometimes the base or exponent of an exponential expression is assigned a specific value, which means we can evaluate the expression at that given value. For instance, if asked to evaluate 𝑎 when 𝑎=2, we substitute for the base 𝑎 to get 2=2×2×2×2=16.

It is also possible to combine terms together to form exponential expressions in more than one variable, such as 𝑎𝑏 or 𝑎×𝑏. Here, we will focus on expressions of this type, with variable bases or exponents, and learn how to evaluate them at given values of the variables.

Example 1: Evaluating a Numerical Exponent Expression Involving Addition

If 𝑎=3 and 𝑏=2, find the value of 𝑎+𝑏.

Answer

The exponential expression 𝑎+𝑏 has two variable bases, 𝑎 and 𝑏, and we must find its value when 𝑎=3 and 𝑏=2. Substituting these values into our expression, we have 3+2=3×3+2×2.

Remembering the correct order of operations, we need to work out the two products individually and then add the results together. Since the square root of a number multiplied by itself gives the original (base) number, we see that 3×3=3 and 2×2=2, so 3×3+2×2=3+2=5.

We deduce that if 𝑎=3 and 𝑏=2, then 𝑎+𝑏=5.

If two exponential expressions have the same base or exponent, then it is possible to combine them together in various ways. We can simplify and evaluate such expressions using the laws of exponents (often referred to as the laws of indices). Since we will need to apply these laws to later examples, we give a brief summary of them below.

Law: Laws of Exponents

For nonzero real numbers 𝑎 and 𝑏 and nonzero integers 𝑚 and 𝑛, these laws of exponents hold: Product:Quotient:Powerofaproduct:Powerofaquotient:Powerofapower:Zeroexponent:Negativeexponent:Rationalexponent:𝑎×𝑎=𝑎𝑎÷𝑎=𝑎𝑎=𝑎(𝑎𝑏)=𝑎𝑏𝑎𝑏=𝑎𝑏(𝑎)=𝑎𝑎=1𝑎=1𝑎𝑎=𝑎

Observe that by using these laws, we can obtain an alternative method of solution for example 1, as follows.

Substituting the values 𝑎=3 and 𝑏=2 into the exponential expression 𝑎+𝑏 gives 3+2.

This time, by the rational exponent law 𝑐=𝑐, we have 𝑐=𝑐=𝑐, which means 3+2=3+2.

Next, the power of a power law (𝑐)=𝑐 implies 𝑐=𝑐=𝑐=𝑐×, so 3+2=3+2=3+2=3+2=5,×× which agrees with our original answer as expected.

Our next example involves another exponential expression with variable bases. In this case, we shall use the above laws to simplify the expression before substituting in the given values of the bases. In particular, these laws offer a means of solving problems that feature negative exponents.

Example 2: Evaluating a Numerical Exponent Expression Involving Subtraction, Multiplication, and Division

If 𝑎=7 and 𝑏=14, find the value of 𝑎×𝑏𝑎𝑏.

Answer

Recall that 𝑎×𝑏𝑎𝑏 is an exponential expression. It has the variable bases 𝑎 and 𝑏, and we need to find its value when 𝑎=7 and 𝑏=14.

To start, we can rewrite the expression as follows: 𝑎×𝑏𝑎𝑏=𝑎×𝑏𝑎×1𝑏.

Notice that the last expression contains two negative exponents. Applying the negative exponent law 𝑐=1𝑐 to the term 𝑎, we find 𝑎=1𝑎. Similarly, applying the rearranged version 1𝑐=𝑐 to the term 1𝑏, we get 1𝑏=𝑏. Our original expression can therefore be rewritten as 𝑎×𝑏1𝑎×𝑏=𝑎×𝑏𝑏𝑎.

Before substituting the given values for 𝑎 and 𝑏 into this expression, we can use the rational index law 𝑐=𝑐 to write 𝑎=7 as 7 and 𝑏=14 as 14, so 𝑎×𝑏𝑏𝑎=7×14147.

Then, we apply the power of a power law (𝑐)=𝑐 in four places, and simplify, to get 7×14147=7×14147=7×14147=7×14147=982=96.××××

Thus, if 𝑎=7 and 𝑏=14, then 𝑎×𝑏𝑎𝑏=96.

Notice how every step of the above example involving the laws of exponents was justified by reference to the relevant law. It is important to give a full explanation of one’s reasoning so that the method is clear.

Example 3: Finding the Numerical Value for an Algebraic Expression at Specific Values Using Laws of Exponents with Negative Exponents

Calculate the numerical value of 54×54 when 𝑥=7 and 𝑦=4.

Answer

In this question, we have the exponential expression 54×54, which has the single numerical base 54 and the variable exponents 𝑥 and 𝑦. We need to calculate its value when 𝑥=7 and 𝑦=4.

As both terms in this product have the same base, they can be combined together using the product law 𝑎×𝑎=𝑎 to get 54×54=54.

Then, we substitute the values 𝑥=7 and 𝑦=4, which gives 54=54.

By the power of a quotient law 𝑎𝑏=𝑎𝑏, this simplifies to 54=5×14.

The next step is to apply the negative exponent law 𝑎=1𝑎 to the term 5, giving 5=15. Similarly, applying the rearranged version 1𝑎=𝑎 to the term 14 gives 14=4. We then have 15×4=45=64125, which is a fraction in its simplest form.

Note that above we could have simplified the exponential expression 54 by another method, as follows. Applying the negative exponent law 𝑎=1𝑎, we have 54=1.

Dividing by a power of a fraction is the same as multiplying by that power turned upside down, that is, multiplying by its reciprocal. Therefore, 1=1×45=45.

Finally, by the power of a quotient law 𝑎𝑏=𝑎𝑏, we have 45=45=64125, as before.

We conclude that when 𝑥=7 and 𝑦=4, the numerical value of 54×54 is 64125.

Our next example involves two exponential equations with numerical bases and variable exponents. This time, we will need to use the laws of exponents to rewrite one base in terms of the other.

Example 4: Evaluating Algebraic Expressions by Solving Exponential Equations Using Laws of Exponents

Given that 2=8=512, determine the value of 𝑥𝑦.

Answer

In this question, we have the two exponential equations 2=5128=512.and

The exponential expressions on the left-hand sides have the numerical bases 2 and 8, respectively, as well as the variable exponents 𝑥 and 𝑦. Our strategy will be to solve these equations for 𝑥 and 𝑦; we can then substitute into the expression 𝑥𝑦 and evaluate it.

In the second equation, notice that the base 8=2×2×2=2, a power of 2. This means we can rewrite the base 8 in terms of the base 2, so the second equation becomes 2=512.

By the power of a power law (𝑎)=𝑎, we have 2=2=2×, so our two exponential equations are now 2=5122=512.and

We see that both equations have the same right-hand side, and that they are identical except in their exponents. This implies that the exponents themselves must be equal, so 𝑥=3𝑦. Therefore, we have reduced the two original equations to a single one: 2=512,𝑥=3𝑦.where

To solve this equation, we need to find the power of 2 that equals 512. We can use a calculator and try different numbers; for instance, 2=2×2×2×2×2=32, which is too small. Examining some higher powers of 2, we have 2=2×2=32×2=64,2=2×2=64×2=128,2=2×2=128×2=256,2=2×2=256×2=512.

Thus, the solution to the exponential equation 2=512 is 𝑥=9. Remember, we also have that 3𝑦=𝑥, so substituting the value of 𝑥 gives 3𝑦=9. Dividing both sides of this equation by 3, we get 𝑦=3.

Lastly, we can determine the value of the algebraic expression 𝑥𝑦 at 𝑥=9 and 𝑦=3: 𝑥𝑦=93=6.

We conclude that when 2=8=512, the value of 𝑥𝑦 is 6.

Finally, the techniques discussed here can be used to answer word problems from other fields. For instance, geometric formulas for area and volume often involve lengths that are squared or cubed (resulting in terms with exponents 2 or 3), so geometry provides a natural context for applying these strategies.

Example 5: Solving a Word Problem by Rearranging and Evaluating an Algebraic Expression

The volume of a right circular cone is given by 𝑉=13𝜋𝑟, where 𝜋=227. If the volume of a right circular cone equals 462 cm3 and the radius 𝑟 of its base is 7 cm, find the height of the cone.

Answer

We have been given the formula for the volume 𝑉 of a cone, 𝑉=13𝜋𝑟.

Looking at the right-hand side, as 13 and 𝜋 are just numbers, this is an exponential expression with the variable bases 𝑟 (the radius) and (the height).

Since we need to find the height of the cone, our first step is to take this formula and rearrange it to make the subject. Multiplying both sides by 3 gives 3𝑉=𝜋𝑟, and dividing through by 𝜋𝑟, we get 3𝑉𝜋𝑟=, which is the same as =3𝑉𝜋𝑟. Now, we are ready to substitute the values 𝑉=462cm, 𝜋=227, and 𝑟=7cm from the question, so =3𝑉𝜋𝑟=3×462×7.

Rewriting the denominator as 227×7×7=22×7×77, we can divide the top and bottom by 7 to simplify it to 22×7. Thus, =3×46222×7=1386154=9.

The radius of the cone is given in centimetres, so its height is also in centimetres; the height of the cone is 9 cm.

Let us finish by recapping some key concepts from this explainer.

Key Points

  • The term “exponential expressions” refers to algebraic expressions of the form 𝑎, where 𝑎 is a real number called the base and 𝑛 is an integer called the exponent (or index, or power). We can evaluate exponential expressions with variable exponents (or bases) at a given value of the exponent (or base).
  • When evaluating more complex exponential expressions, we can use the laws of exponents to simplify them as much as possible.
    For nonzero real numbers 𝑎 and 𝑏 and nonzero integers 𝑚 and 𝑛, these laws hold: Product:Quotient:Powerofaproduct:Powerofaquotient:Powerofapower:Zeroexponent:Negativeexponent:Rationalexponent:𝑎×𝑎=𝑎𝑎÷𝑎=𝑎𝑎=𝑎(𝑎𝑏)=𝑎𝑏𝑎𝑏=𝑎𝑏(𝑎)=𝑎𝑎=1𝑎=1𝑎𝑎=𝑎
  • These techniques can be applied to solve geometric and real-world word problems that require evaluating exponential expressions with variable exponents or bases.

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