Question Video: Identifying the Power of an 𝑛th Root Irrational Number That Is a Rational Number Mathematics

Let π‘₯ = the fifth root of (βˆ›(7)/∜(2)). Which of the following numbers is rational? [A] π‘₯Β³ [B] π‘₯⁴ [C] π‘₯ΒΉΒ² [D] π‘₯¹⁡ [E] π‘₯⁢⁰

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Video Transcript

Let π‘₯ be equal to the fifth root of the cubed root of seven over the fourth root of two. Which of the following numbers is rational? (A) π‘₯ cubed, (B) π‘₯ to the fourth power, (C) π‘₯ to the 12th power, (D) π‘₯ to the 15th power, or (E) π‘₯ to the 60th power.

Before we start, there are two things we should think about, the first of them being what is a rational number. A rational number is a number that can be written in the form 𝑝 over π‘ž, where 𝑝 and π‘ž are integers and π‘ž is not equal to zero. Another way to say this is that a rational number can be made by dividing two integers.

Now, the second thing we might want to think about is seeing if we can simplify what π‘₯ is equal to here. To do that, let’s think about our exponent rule that tells us that the 𝑛th root of π‘₯ is equal to π‘₯ to the one over 𝑛 power. And that means we can rewrite the fifth root as the one-fifth power. And we wanna repeat this process with the two roots inside the parentheses, which means we’ll have seven to the one-third power over two to the one-fourth power, all taken to the one-fifth power.

And one further step we can do to simplify is remember the power of a power rule that tells us π‘₯ to the π‘Ž power to the 𝑏 power is equal to π‘₯ to the π‘Ž times 𝑏 power. And this means we can multiply one-third by one-fifth. When we do that, we multiply the numerators, one times one is one, then multiply the denominators, three times five is 15. We’ll have seven to the 15th power.

And then we’ll need to multiply one-fourth times one-fifth to find the power of our denominator. One-fourth times one-fifth is one twentieth. And so we’re saying that π‘₯ is equal to seven to the one fifteenth power over two to the one twentieth power. To find a rational value, we need the numerator and the denominator here to be an integer. So let’s consider some of our options.

If we cube π‘₯, we’ll have seven to the one fifteenth power over two to the one twentieth power cubed. And if we were going to take a power of a power, we would multiply these powers together. We would have π‘₯ cubed being equal to seven to the three fifteenth power over two to the three twentieths power.

Now, we could plug this value into our calculator to see what would happen. When we do that, we get 1.3304 continuing. This is an irrational value. And so maybe we should consider what kind of powers here would give us integers. If our exponents were whole numbers, if we were taking seven to an integer power, the outcome would be an integer. And that means we want to take both two and seven to some power of an integer. We need to multiply one fifteenth and one twentieth by some number that produces an integer for both of these values.

For example, if we took π‘₯ to the 15th power, then we would have seven to the first power. But we would still have a fraction in our denominator. But that does get us a bit closer. From our list, we can take π‘₯ to the 60th power, which would be seven to the 60 over 15 power and two to the 60 over 20 power. 60 divided by 15 is four. 60 divided by 20 is three. π‘₯ to the fourth power will be an integer, and two cubed will also be an integer. An integer divided by an integer is a rational number. And so we can say that π‘₯ to the 60th power will be rational.

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