Video: Simplifying Expressions: Negative and Fractional Exponents

In this video, we will learn how to use the rules of negative and fractional indices to solve problems.

15:13

Video Transcript

In this video, we’re going to look at simplifying expressions, particularly expressions that have negative and fractional exponents in them.

First, we can address what simplifying an expression is. By simplifying, we mean writing an expression in the most compact or efficient manner without changing the value of that expression. The process in simplifying involves removing parentheses by multiplying factors, combining like terms, or, like we’ll be considering today, simplifying exponents. We might also wonder why we would do this simplification process.

Simplifying makes expressions easier to read and understand. Additionally, it reduces possible errors in calculating. This is true because after you’ve simplified an expression, there should be less calculations to do than in an unsimplified form. To do this simplifying when our expression has exponents, we’ll need to remember some exponent rules.

We have the rule for multiplying exponents together. π‘₯ to the π‘Ž power times π‘₯ to the 𝑏 power equals π‘₯ to the π‘Ž plus 𝑏 power. If we have π‘₯ squared times π‘₯ cubed, that will be π‘₯ to the two plus three power, which is π‘₯ to the fifth. Next, we have our quotient rule. π‘₯ to the π‘Ž power divided by π‘₯ to the 𝑏 power equals π‘₯ to the π‘Ž minus 𝑏 power. Here, we have π‘₯ to the fifth power divided by π‘₯ squared, which will be equal to π‘₯ to the five minus two power, π‘₯ cubed.

The next rule is called a power of a power. π‘₯ to the π‘Ž power to the 𝑏 power is equal to π‘₯ to the π‘Ž times 𝑏 power. If we have π‘₯ cubed and we’re taking that to the fourth power, it will be π‘₯ to the three times four power, π‘₯ to the 12th power. Another rule, π‘₯ divided by 𝑦 to the π‘Ž power is equal to π‘₯ to the π‘Ž power over 𝑦 to the π‘Ž power, which is a very similar rule to π‘₯ times 𝑦 to the π‘Ž power is equal to π‘₯ to the π‘Ž power times 𝑦 to the π‘Ž power.

And then, we have our negative exponent rule. π‘₯ to the negative π‘Ž power is equal to one over π‘₯ to the positive π‘Ž power. π‘₯ to the negative fifth power is equal to one over π‘₯ to the fifth power. By the same rule, we can say one over π‘₯ to the negative three power is equal to π‘₯ to the positive three power. And that means when we have an exponent with a negative in the numerator and an exponent with a negative in the denominator. The π‘₯ to the negative four power moves into the denominator and becomes π‘₯ to the positive four power. The 𝑦 to the negative three power in the denominator moves to the numerator and becomes positive.

And before we move on, we should note that all five of the rules above this are also true when the exponent value is negative. For example, if we have π‘₯ to the negative three power being multiplied by π‘₯ to the negative five power, it will be equal to π‘₯ to the negative three plus negative five power, π‘₯ to the negative eight.

We’re nearly ready to look at some examples, but first we need to consider fractional exponents. Fractional exponents, sometimes called rational exponents, represent certain roots. What we’re saying here is that 𝑛th root of π‘₯ is equal to π‘₯ to the one over 𝑛 power. By this rule, the square root of π‘₯ is equal to π‘₯ to the one-half power. We know that this is the square because when this radical symbol does not have a value in front of it, it’s the square root.

On the other hand, the cube root of π‘₯ would be equal to π‘₯ to the one-third power. This fractional exponent occurs in one more form. And that’s when we have π‘₯ to the π‘š power and we’re taking the 𝑛th root of that. That will be equal to π‘₯ to the π‘š over 𝑛 power. If we have the cube root of π‘₯ squared, that is π‘₯ to the two-third power. The denominator of the exponent comes from the root, and the numerator comes from the power.

Another thing we need to say about these kinds of fractional exponents is that it’s also true if you first take the cube root of π‘₯ and then you square that value. The way that it’s written here, you would take π‘₯ squared first, and then you would do the cube root. In this format, you would take the cube root and then you would square that value. Both of these would give you the same numerical result. And they’re both equal to π‘₯ to the two-thirds power. Now, we’re ready to consider some examples.

Which of the following is equal to three-fifths to the negative six power times three-fifths to the negative three power all over three-fifths to the eight power? (A) Three-fifths to the 11th power, (B) three-fifths to the negative one power, (C) three-fifths to the negative 11 power, (D) three-fifths to the negative 17th power, or (E) three-fifths to the negative 25 power.

Here’s our expression. In our numerator, we have two exponents with the same base being multiplied together. And we know when that happens, we can simplify by adding the two powers together. This is true even though both of these values are negative. To simplify then, we’ll have three-fifths to the negative six plus negative three power, which is three-fifths to the negative ninth power. We bring across the denominator. And then we see we’re dividing an exponent by another exponent with the same base.

And to do that, we know that we can subtract the power values. This is true even though one of the powers is negative. It would be three-fifths to the negative nine minus eight power, three-fifths to the negative 17th power. Now, it’s possible to distribute this negative 17 to simplify even further. However, our question is just asking to find an equivalent expression, which we have, three-fifths to the negative 17th power, which is the final answer.

In our next example, we’ll need to simplify with these fractional powers.

Simplify 16 to the five-fourth power over 16 to the one-half power.

We have a few choices about the order in which we choose to solve this. Because both of these values have the same base, we could subtract their powers, which would make it 16 to the five-fourth minus one-half power. Now, one-half is two-fourths, which means we would have five-fourths minus two-fourths, which is 16 to the three-fourth power. This fractional power tells us we’re taking the fourth root of 16 cubed.

Because of our power of a power rule, we can say 16th to the one-fourth power to the third power or we could say 16 cubed to the one-fourth power. 16 to the one-fourth power cubed would be my preferred option because 16 to the one-fourth power is two, whereas 16 cubed is 4096. From there, two cubed equals eight. And to take the fourth root of 4096, you would need a calculator. But once you put it in the calculator, it would tell you that the fourth root of 4096 is eight.

By taking the fourth root first and then cubing the answer, we were able to keep the values bit smaller than if you did it the other way around. But both ways show us that this expression is equal to eight.

Here’s our next example.

Calculate the square root of one-fourth to the fifth power times one-fourth squared.

Looking at this expression, we notice that inside the radical, we’re dealing with exponents that have the same base. We can combine these values by saying one-fourth to the five plus two power, one-fourth to the seventh power. And then, we can take this radical, and we can rewrite it as a fractional exponent so that we have one-fourth to the seven-halves power. And then, we have a choice. We can either take one-fourth to the one-half power and then take that value to the seventh power. Or we can say one-fourth to the seventh power, find out what that is, and then take the square root of that value.

Taking the square root first usually means the numbers that you’re working with are a bit simpler. We can distribute that one-half power over the one and the four. The square root of one is one, and the square root of four is two. So, we now have one-half to the seventh power. When we distribute that seventh power across the one and the two, one to the seventh power is one, and two to the seventh power is 128. The value of this expression is then one out of 128.

Here’s another example.

Which of the following expressions has the same value as negative two to the fifth power to the negative third power? (A) Negative two squared, (B) negative two to the eighth power, (C) negative two to the 15th power, (D) negative one over two to the eighth power, or (E) negative one over two to the 15th power.

We have the expression negative two to the fifth power to the negative three power. Since we’re taking a power of a power, we can go ahead and multiply these two indexes together. That would be five times negative three, which is negative 15. Be careful here; one of the answer choices is negative two to the positive 15th power. And that’s not what we see here. So, we might think π‘₯ to the negative π‘Ž power equals one over π‘₯ to the π‘Ž power, which gives us one over negative two to the 15th power, but still not exactly what we’re looking for.

One thing we can say about negative two is that it has factors negative one and two. And that means if we break these factors apart, we’re able to distribute that power 15. This gives us one over negative one to the 15th power times two to the 15th power. Because 15 is an odd number, negative one to the 15th power is negative one. We know this because negative one times negative one is positive one, but negative one times negative one times negative one is negative one.

We take that multiplied by negative one and make this fraction negative so that the simplified form is negative one over two to the 15th power, which is option (E).

In this example, we have two different variables and negative exponents.

Simplify π‘š over 𝑛 to the negative one power all taken to the negative third power times two times π‘š to the negative two power over 𝑛 to the negative two power all taken to the negative three power.

That was quite a mouthful. Let’s go ahead and copy down this expression. The first thing that we see is that we have a power of a power. Both of these fractions are being taken to the negative three power. On top of that, we know that when we’re dividing values and they’re being taken to the same power, we can distribute that power. Which means for this first fraction, I can write π‘š to the negative three power over 𝑛 to the positive three power.

This is positive three because negative one times negative three is positive three. For the second fraction, we have two to the negative three power and then π‘š to the negative two times negative three power, which will be π‘š to the positive six power. And then, in the denominator, 𝑛 to the negative two times negative three, 𝑛 to the positive six power.

We know when we’re multiplying fractions, we can multiply the numerators and multiply the denominators. We can multiply π‘š to the negative three power times π‘š to the positive six power, which will be π‘š to the negative three plus six power, π‘š cubed. And we still have that two to the negative three power in the numerator. In the denominator, we have 𝑛 cubed times 𝑛 to the sixth power, which will be 𝑛 to the three plus sixth power, 𝑛 to the ninth power.

We nearly have a simplified form. We can’t simplify π‘š cubed any further, or 𝑛 to the ninth. However, we still have this two to the negative three power. And that means we need to bring that value into the denominator, which gives us π‘š cubed over two cubed times 𝑛 to the ninth. And we know that two cubed is eight, which makes the simplified form of this expression π‘š cubed over eight times 𝑛 to the ninth power.

In our final example, we’re going to consider something we haven’t seen before, a mixed number as a base for an exponent.

Which of the following is equal to one and three-fifths squared times one and three-fifths to the negative three power? (A) Five-eighths, (B) 25 over 64, (C) eight-thirds, (D) negative five-eighths, or (E) negative eight-fifths.

We copy down our expression. Before we get started, we need to make one very important clarification. We have the rule that π‘₯ times 𝑦 squared is equal to π‘₯ squared 𝑦 squared. You might be tempted to write one squared times three-fifths squared. This is not true. And that’s because the mixed number one and three-fifths represents one plus three-fifths. And it does not represent one times three-fifths. And since we can’t do that kind of distribution, we need a new strategy to simplify.

We’ll need to convert these mixed numbers into improper fractions. The improper fraction will be one times five plus three, which is eight, over the original denominator of five. Both of these values are the same mixed number, so they’re both the eight-fifths as an improper fraction. Once we’re to this point, since these exponents have the same base, we can add their powers. Two plus negative three is negative one. And from there, we are able to distribute that power so that we have eight to the negative one power over five to the negative one power.

Since we have a negative power in the numerator and a negative power in the denominator, we can flip them. Five to the first power is five. Eight to the first power is eight. The equivalent expression is five-eighths, which is option (A). The key to solving this one was recognizing that you needed to change the format of these mixed numbers before you went about solving.

From here, we’re ready to summarize the key points. Simplifying expressions allows to write expressions in the most compact and efficient manner without changing the value of the expression. And when these expressions have negative and fractional exponents, we use these rules to simplify the terms.

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