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Lesson Video: Simplifying Exponential Expressions with Rational Exponents Mathematics • 9th Grade

In this video, we will learn how to perform operations and simplifications with expressions that involve rational exponents.

18:06

Video Transcript

In this video, we’ll learn how to perform operations and simplifications with expressions that involve rational exponents.

You should already be familiar with the laws of exponents for integer powers and 𝑛th roots. Recall that a rational number is another name for a fraction. Both the numerator and denominator are integers, that is, whole numbers, and 𝑛, the denominator, is nonzero. So, a rational exponent is simply a power or exponent that is a fraction. So, if we have a number, a base π‘Ž, we can raise this to a rational power. If the numerator of our rational exponent is equal to one, then we have our base π‘Ž raised to the power one over 𝑛.

And we say that a fractional exponent one over 𝑛, where 𝑛 is an integer, can be expressed as the 𝑛th root of π‘Ž. For example, eight raised to the power one over three is the cubed root of eight. And that’s equal to two since two multiplied by two multiplied by two is equal to eight. That is, two is the number we cube to get to eight. So, two is the cubed root of eight.

Now, before we start looking at some examples of simplifying exponential expressions with rational exponents, let’s remind ourselves of the laws of exponents that apply to both rational and integer exponents. We recall that we have two product rules for exponents. The first is π‘Ž raised to the power 𝑛 multiplied by π‘Ž raised to the power π‘š is equal to π‘Ž raised to the power π‘š plus 𝑛. So, we sum our powers. And π‘Ž raised to the power 𝑛 multiplied by 𝑏 raised to the power 𝑛 is π‘Ž multiplied by 𝑏 raised to the power 𝑛. Our rule for negative exponents is π‘Ž raised to the power negative 𝑛 is one over π‘Ž to the power 𝑛.

Our quotient rules are π‘Ž raised to the power 𝑛 over π‘Ž raised to the power π‘š is equal to π‘Ž raised to the power 𝑛 minus π‘š. So, we subtract the power. And π‘Ž over 𝑏 raised to the power 𝑛 is equal to π‘Ž raised to the power 𝑛 multiplied by 𝑏 raised to the power negative 𝑛.

Our power rules are π‘Ž raised to the power 𝑛 to the power π‘š is equal to π‘Ž raised to the power 𝑛 multiplied by π‘š. So, we multiply our powers. And π‘Ž raised to the power π‘š over 𝑛, that is, a rational power or rational exponent, is equal to the 𝑛th root of π‘Ž raised to the power π‘š. We note also that π‘Ž raised to the power zero is equal to one for any nonzero π‘Ž. And one raised to the power 𝑝 is equal to one if 𝑝 is a rational number.

So, let’s consider our first example, where we use our knowledge of exponents to simplify an expression with a rational exponent.

Simplify negative 64π‘Ž raised to the power 12 multiplied by 𝑏 raised to the power 18 to the power one over six, where π‘Ž and 𝑏 are positive constants.

We’re given an expression raised to a rational exponent. And to simplify this, we’re going to make use of the power law for exponents, which says that π‘Ž raised to the power 𝑛 to the power π‘š is equal to π‘Ž raised to the power 𝑛 multiplied by π‘š. First, however, we note that we can decompose 64 into its prime factors. In fact, 64 is two raised to the power six. So, now, separating out our factors, we have negative two raised to the power six raised to the power one over six multiplied by π‘Ž raised to the power 12 to the power one over six multiplied by 𝑏 raised to the power 18 to the power one over six.

And now, using our power law, we can write our expression as negative two raised to the power six multiplied by one over six multiplied by π‘Ž to the power 12 multiplied by one over six times 𝑏 to the power 18 multiplied by one over six. And since six multiplied by one-sixth is equal to one, 12 multiplied by one-sixth is equal to two, and 18 multiplied by one-sixth is equal to three, we have negative two raised to the power one multiplied by π‘Ž squared multiplied by 𝑏 cubed. And since any number to the power one is itself, we have that the expression negative 64 multiplied by π‘Ž to the power 12 multiplied by 𝑏 raised to the power 18 all to the power one over six is equal to negative two π‘Ž squared 𝑏 cubed.

Now, let’s try another slightly more complex example, where we simplify an expression with a rational exponent.

Expand the negative of π‘Ž cubed plus one over π‘Ž squared all raised to the power 10 all raised to the power one over five, where π‘Ž is a real constant.

We’re given an expression in terms of a real constant π‘Ž, which is raised to the power 10. And we have the negative of this expression, which is all raised to a rational power one over five. And to simplify this expression or expand it, we’re going to use the power law for exponents, which tells us that π‘Ž raised to the power 𝑛 to the power π‘š is equal to π‘Ž raised to the power 𝑛 multiplied by π‘š. We note first, in fact, that our negative sign means that we’re multiplying inside our parentheses by negative one. And so, using our power law, we have negative one raised to the power one over five multiplied by π‘Ž cubed plus one over π‘Ž squared to the power 10 all raised to the power one over five.

Now, noting that multiplying five instances of negative one together gives us negative one, then we can say that the fifth root of negative one is also negative one. So now with this, together with our power rule for exponents, we have negative π‘Ž cubed plus one over π‘Ž squared raised to the power 10 multiplied by one over five. Our exponent 10 multiplied by one over five is equal to 10 over five, and that’s equal to two. And making some room, this leaves us with negative π‘Ž cubed plus one over π‘Ž squared all squared.

And now, squaring our expression, we have negative π‘Ž raised to the power six plus two π‘Ž cubed over π‘Ž squared plus one over π‘Ž to the power four. Dividing the numerator and denominator of our second term by π‘Ž squared gives us two π‘Ž. And multiplying through by negative one, we have negative π‘Ž raised to the power six minus two π‘Ž minus one over π‘Ž raised to the power four. Hence, the expression negative π‘Ž cubed plus one over π‘Ž squared to the power 10 all raised to the one over five is equal to negative π‘Ž raised to the power six minus two π‘Ž minus one over π‘Ž raised to the power four.

Before our next example, let’s recap again what we know about rational exponents. Fractional or rational exponents are exponents with integer numerator and denominator, where the denominator 𝑛 is nonzero. And using the power rule for exponents, we can say that π‘Ž raised to the power π‘š over 𝑛 is equal to π‘Ž raised to the power one over 𝑛 to the power π‘š. And π‘Ž raised to the power one over 𝑛 is the 𝑛th root of π‘Ž. So, in fact, we have the 𝑛th root of π‘Ž raised to the power π‘š.

Equivalently, we can take the rational exponent on the outside. So, we have π‘Ž raised to the power π‘š over 𝑛 is equal to π‘Ž raised to the power π‘š all raised to the power one over 𝑛. And that’s equal to the 𝑛th root of π‘Ž raised to the power π‘š. And so we have π‘Ž raised to the power π‘š over 𝑛, the rational power, is equal to the 𝑛th root of π‘Ž all to the power π‘š. And that’s equal to the 𝑛th root of π‘Ž to the power π‘š.

We’ll use this in our next example together with prime factorization to break down our bases. This can help us to simplify some complicated numerical expressions with different bases.

Simplify 36 to the power one over four multiplied by 21 squared multiplied by eight raised to the power one over five all divided by 486 raised to the power one over 10 multiplied by 42 cubed.

We’re given a rational expression to simplify, where the numerator and denominator comprise integers raised to rational powers, that is, where the exponents are fractions. To do this, we’ll use the laws of exponents as shown. But first we’ll decompose our bases, that’s 36, 21, eight, 486, and 42, into their prime factorizations. We know that 36 is three squared multiplied by two squared, 21 is three times seven, eight is two cubed, 486 is two multiplied by three raised to the power five, and 42 is two multiplied by three multiplied by seven.

Our expression is now three squared multiplied by two squared to the power one over four multiplied by three times seven all squared multiplied by two cubed to the power one over five all divided by two multiplied by three to the power five all to the power one over 10 multiplied by two times three times seven all cubed. Using our rule number one, we can bring our exponents inside our parentheses. And we can also use rule four at the same time; that’s the power rule. And noting that two over four is equal to one-half and also that five over 10 is equal to one-half, and collecting our like bases in both numerator and denominator, we have three raised to the power of a half multiplied by three squared multiplied by two raised to the power of a half multiplied by two to the power of three over five multiplied by seven squared divided by three raised to the power of a half multiplied by three cubed multiplied by two raised to the power one over 10 multiplied by two cubed multiplied by seven cubed.

And at this point, we can use our power rule three and our negative exponent rule number five to collect our exponents for each of our bases so that we have three raised to the power of a half plus two minus one-half minus three multiplied by two raised to the power of a half plus three-fifths minus one-tenth minus three multiplied by seven raised to the power two minus three. Evaluating one-half plus two minus one-half minus three, this gives us negative one. And similarly, for our power of two, we have one-half plus three-fifths minus one-tenth minus three is equal to negative two. And two minus three for seven is negative one. And so we have three raised to the power negative one multiplied by two to the power negative two multiplied by seven raised to the power negative one.

And again, using our rule number five, we have one over three multiplied by two squared multiplied by seven. And evaluating this gives us one over 84. And hence, our expression 36 to the power one over four multiplied by 21 squared multiplied by eight to the power one over five all divided by 486 to the power one over 10 multiplied by 42 cubed simplifies to one over 84.

In the case where our bases are decimal, we would write this base as a fraction and then decompose the numerator and denominator of this fraction into their prime factors. So, let’s look at an example of this.

Simplify 0.25 raised to the power three over two multiplied by 1.8 squared all over eight raised to the power two over three.

In this example, we want to simplify a numerical expression which combines decimals and rational exponents. Let’s begin by noting some of the laws of exponents that we’ll need. Before we use these laws, however, let’s write each of the decimal bases as a fraction. We can then express these in terms of their prime factorizations. Our bases are 0.25, 1.8, and eight. And we know that 0.25 is one-quarter, and that’s equal to one over two squared. 1.8 is equal to nine over five, and that’s three squared over five. And we know that eight is equal to two cubed. Our expression can then be written as one over two squared raised to the power three over two multiplied by three squared over five all squared all over two cubed raised to the power two over three.

Now, using rules two and rule four, our expression becomes one over two raised to the power two multiplied by three over two all multiplied by three to the power two times two all over five squared. And that’s all over two raised to the power three multiplied by two over three. And evaluating our exponents, we have two multiplied by three over two is equal to three, two squared is four, and three multiplied by two over three is equal to two. What we now have then is three to the power four divided by two cubed multiplied by two squared multiplied by five squared. We can sum our powers of two using rule number three so that we have three raised to the power four divided by two to the power five multiplied by five squared. And evaluating this gives us 81 over 800. Our expression then 0.25 raised to the power three over two multiplied by 1.8 squared all divided by eight to the power two over three simplifies to 81 over 800.

In our final example, we simplify an expression involving rational and negative variable exponents.

Simplify 25 raised to the power three over two π‘₯ multiplied by eight raised to the power π‘₯ minus five over three all divided by 100 raised to the power three over two π‘₯ multiplied by the square root of 100.

The expression we want to simplify in this example involves rational and negative exponents. So, let’s make a note of the laws of exponents we’re going to need to use. Now, assuming that π‘₯ is a real number, let’s begin by looking at the exponents in each of the factors. In our first factor, that’s 25 raised to the power three over two π‘₯. We can use rule number four to bring in the power of a half. And since we know that 25 raised to the power of a half is the square root of 25, and that’s equal to five, our first factor is five raised to the power three π‘₯.

Our second factor in the numerator is eight raised to the power π‘₯ minus five over three. And we can use the product rule number three to break this down to eight to the power of π‘₯ multiplied by eight to the power negative five over three. Next, using rule number four on the second part of this, we have eight raised to the power π‘₯ multiplied by eight to the power one over three to the power negative five. Now we know that eight raised to the power one over three is equal to two because eight to the one over three is the cubed root of eight. And since eight is equal to two multiplied by two times two, that’s equal to two cubed. The cubed root of eight is two. So, in fact, eight raised to the power π‘₯ minus five over three is eight to the power π‘₯ multiplied by two to the power negative five.

And now using rule number four again on 100 to the power three over two π‘₯, we have 100 to the power of a half to the power three π‘₯. And we know that 100 to the power of a half, that is, the square root of 100, is equal to 10. So that 100 to the power three over two π‘₯ is equal to 10 to the power three π‘₯. So, now making some room, we have 25 raised to the power three over two π‘₯ is equal to five to the power three π‘₯. Eight to the power π‘₯ minus five over three is eight to the power π‘₯ multiplied by two to the negative five. And 100 to the power three over two π‘₯ is 10 raised to the power three π‘₯. And our final factor in the denominator is the square root of 100, which is 10.

Our expression now becomes five raised to the power three π‘₯ multiplied by eight to the power π‘₯ multiplied by two to the negative five all over 10 to the power three π‘₯ multiplied by 10. So, now, decomposing our bases of eight and 10 into prime factors, we know that eight is equal to two cubed and 10 is two times five. So, now, we have five to the power three π‘₯ multiplied by two cubed to the power π‘₯ multiplied by two to the negative five power all over two times five to the power three π‘₯ multiplied by two times five.

And now using rules one and four, we can bring our powers in. We can divide both our numerator and denominator by five raised to the power three π‘₯ and also by two raised to the power three π‘₯. And we’re left with two raised to the power negative five divided by two multiplied by five. Using rule number five, we see the two to the negative five is one over two to the power five. And so we have one over two multiplied by two to the power five multiplied by five. And by rule number three, this is one over two to the power six multiplied by five. And since two to the power six is 64, we have one over 64 multiplied by five, which is one over 320.

The expression 25 raised to the power three over two π‘₯ multiplied by eight to the power π‘₯ minus five over three all divided by 100 to the power three over two π‘₯ multiplied by the square root of 100 simplifies to one over 320.

Now, let’s complete this video by recapping some of the key points we’ve covered. We’ve used the laws of exponents for the bases π‘Ž and 𝑏 and rational exponents 𝑛, π‘š. We have the product rules π‘Ž raised to the power 𝑛 multiplied by π‘Ž to the power π‘š is equal to π‘Ž raised to the power 𝑛 plus π‘š; that’s the sum of the powers. π‘Ž to the power 𝑛 multiplied by 𝑏 to the power 𝑛 is π‘Ž multiplied by 𝑏 all to the power 𝑛. We have the negative exponent rule, which is π‘Ž raised to the power negative 𝑛 is one over π‘Ž to the power 𝑛. Our quotient rules tell us that π‘Ž raised to the power 𝑛 over π‘Ž to the power π‘š is equal to π‘Ž to the power 𝑛 minus π‘š. And π‘Ž to the power 𝑛 over 𝑏 to the power 𝑛 is π‘Ž over 𝑏 all to the power 𝑛.

And finally, our power rules π‘Ž to the power 𝑛 all to the power π‘š is π‘Ž to the power 𝑛 multiplied by π‘š. And π‘Ž to the rational power 𝑛 over π‘š is the π‘šth root of π‘Ž raised to the power 𝑛. We note also that π‘Ž raised to the power zero is equal to one for any π‘Ž not equal to zero and that one raised to the power 𝑝 is equal to one for a rational number 𝑝. When simplifying numerical expressions with different bases or exponents, we can use prime factorization to simplify the bases and collect like bases. We then use the laws of exponents to simplify our expressions.

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