Lesson Video: Powers and Exponents for Rational Numbers | Nagwa Lesson Video: Powers and Exponents for Rational Numbers | Nagwa

Lesson Video: Powers and Exponents for Rational Numbers Mathematics • First Year of Preparatory School

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In this video, we will learn how to identify the base and exponent in power formulas, write them in exponential and expanded forms, and evaluate simple powers.

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Video Transcript

In this video, we will learn how to identify the base and exponent in power formulas, write them in exponential and expanded forms, and evaluate simple powers.

We begin by recalling that we can represent repeated multiplication as a power. For example, two to the fifth power is defined as the product of five twos as shown. We call two the base and five the exponent. We can extend this definition to general rational bases. In this case, if 𝑛 is a positive integer and π‘Ž over 𝑏 is a rational number, then π‘Ž over 𝑏 to the 𝑛th power will be the product of 𝑛 lots of π‘Ž over 𝑏. For example, we can evaluate one-half to the third power, or one-half cubed, by multiplying one-half by one-half and one-half again. Recalling our rules for multiplying fractions, we simply multiply the numerators and denominators separately. This gives us one over eight, or one-eighth.

We can also follow the same process in reverse. Let’s imagine we want to write the fraction 27 over eight in exponential form. We begin by factoring the numerator and denominator into primes as follows. 27 is equal to three multiplied by three multiplied by three, and eight is equal to two multiplied by two multiplied by two. We can then split the multiplication, giving us three over two multiplied by three over two multiplied by three over two. As such, 27 over eight is the product of three lots of three over two and can therefore be written as three over two cubed.

Let’s now consider some examples involving the powers of rational numbers.

What terminology do we use to describe the half in the expression one-half to the fifth power and the five in the expression one-half to the fifth power?

We begin by recalling that an expression of the form π‘Ž to the 𝑛th power is called an exponential expression or the 𝑛th power of π‘Ž. We call π‘Ž the base of the expression and 𝑛 the exponent or power. In this question, we have the expression one-half to the fifth power. The one-half is the number that is being taken to a power, and the five is the power itself. We can therefore conclude that in the expression one-half to the fifth power, one-half is called the base of the expression and five is called the exponent of the expression.

In our next example, we will simplify an expression by rewriting it in exponential form.

What is four over 11 multiplied by four over 11 multiplied by four over 11 multiplied by four over 11 multiplied by four over 11 multiplied by four over 11 multiplied by four over 11? Is it option (A) four elevenths to the power of negative seven? (B) Four elevenths to the power of seven. (C) Four elevenths to the power of nine. (D) Seven elevenths to the power of four. Or (E) twenty-eight elevenths to the power of seven.

We could evaluate this expression by multiplying all of the numerators and all of the denominators. This would give us the following expression which we could calculate with or without a calculator. However, the five options in this question are given as powers. This means that instead of evaluating the expression, we can simplify by recalling that repeated multiplication can be written in exponential form. In particular, in this question, we are multiplying seven lots of four elevenths. We know that the product of seven lots of four elevenths can be written by raising four elevenths to an exponent of seven. And as such, we can conclude that the correct answer is option (B). The expression in the question is equivalent to four elevenths to the power of seven.

We will now use the information we have seen so far to define a key property of the powers of rational numbers. Since a positive integer power of a rational base is defined by repeated multiplication, we can show that if 𝑛 is a positive integer and π‘Ž over 𝑏 is a rational number, then π‘Ž over 𝑏 to the 𝑛th power is equal to π‘Ž to the 𝑛th power divided by 𝑏 to the 𝑛th power. In other words, we can raise the numerator and denominator to the power separately.

Let’s now consider an example where we can apply this.

Find the value of negative six-fifths cubed, giving your answer in its simplest form.

We can evaluate the expression in this question in two ways. Firstly, we recall that negative six over five cubed can be written as repeated multiplication. It is the product of three lots of negative six over five as shown. We can then multiply the numerators and denominators separately, giving us negative six multiplied by negative six multiplied by negative six over five multiplied by five multiplied by five. Multiplying three negative numbers gives a negative answer, so the numerator is negative 216. The denominator simplifies to 125. The value of negative six over five cubed is negative 216 over 125.

We can also evaluate the expression by recalling the general result for powers of rational numbers. If 𝑛 is a positive integer and π‘Ž over 𝑏 is a rational number, then π‘Ž over 𝑏 to the 𝑛th power is equal to π‘Ž to the 𝑛th power divided by 𝑏 to the 𝑛th power. This means that we can rewrite negative six over five cubed as negative six cubed over five cubed, which once again simplifies to negative 216 over 125.

In our next example, we will consider a problem in context.

Find an expression for the volume of the given cube whose side lengths are two π‘₯ over five.

We begin by recalling that the volume of a cube is given by the cube of its side length. This means that if a cube has side length 𝑙, then its volume is 𝑙 multiplied by 𝑙 multiplied by 𝑙, which is 𝑙 cubed. In this question, we are told that the side length is two π‘₯ over five. This means that its volume is given by the expression two π‘₯ over five cubed. Recalling that for any rational number π‘Ž over 𝑏 and any integer 𝑛 π‘Ž over 𝑏 to the 𝑛th power is equal to π‘Ž to the 𝑛th power divided by 𝑏 to the 𝑛th power, our expression simplifies to two π‘₯ cubed over five cubed.

Next, we recall that to multiply monomials, we multiply the coefficients and add the powers of the shared variables. This means that two π‘₯ multiplied by two π‘₯ multiplied by two π‘₯ is equal to eight π‘₯ cubed. And since five cubed is 125, our expression simplifies to eight π‘₯ cubed over 125. This is an expression for the volume of a cube whose side lengths are two π‘₯ over five.

We will now consider one final example where we need to evaluate an algebraic expression using the results for the powers of rational numbers.

If π‘₯ is equal to three over two and 𝑦 is equal to negative four over five, find the value of π‘₯ squared 𝑦 minus π‘₯𝑦 cubed, giving your answer as a fraction in simplest form.

We will begin this question by substituting the values of π‘₯ and 𝑦 into our expression. π‘₯ squared 𝑦 minus π‘₯𝑦 cubed is therefore equal to three over two squared multiplied by negative four over five minus three over two multiplied by negative four over five cubed. Our next step is to evaluate the powers by recalling that if 𝑛 is a positive integer and π‘Ž over 𝑏 is a rational number, then π‘Ž over 𝑏 to the 𝑛th power is equal to π‘Ž to the 𝑛th power over 𝑏 to the 𝑛th power. This means that three over two squared is equal to three squared over two squared, which is equal to nine over four. In the same way, negative four over five cubed is equal to negative four cubed over five cubed. And this is equal to negative 64 over 125. Our expression therefore simplifies to nine over four multiplied by negative four over five minus three over two multiplied by negative 64 over 125.

We can simplify the first part of our expression by dividing the numerator and denominator by four. In the same way, we can divide the numerator and denominator of the second part of our expression by two. And our expression simplifies to negative nine-fifths minus negative 96 over 125. This in turn simplifies to negative nine over five plus 96 over 125. Finally, in order to add fractions, we must have the same denominators, so we multiply the numerator and denominator of our first fraction by 25. The expression becomes negative 225 over 125 plus 96 over 125. Finally, we add the numerators, giving us negative 129 over 125. Since there are no shared factors apart from one, this is the value of π‘₯ squared 𝑦 minus π‘₯𝑦 cubed as a fraction in its simplest form.

We will now finish this video by recapping the key points. We saw that in an expression of the form 𝑏 to the 𝑛th power, we call 𝑏 the base and 𝑛 the power or exponent. We define positive integer powers by repeated multiplication, known as expanded form. In general, if π‘Ž over 𝑏 is a rational number and 𝑛 is a positive integer, then π‘Ž over 𝑏 to the 𝑛th power is a product of 𝑛 lots of π‘Ž over 𝑏 as shown. Finally, we saw that we can evaluate the power of a rational number by evaluating the power of the numerator and denominator separately. If 𝑛 is a positive integer and π‘Ž over 𝑏 is a rational number, then π‘Ž over 𝑏 to the 𝑛th power is equal to π‘Ž to the 𝑛th power divided by 𝑏 to the 𝑛th power.

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