Question

question 38 of 41
which of the following is equivalent to a real number?

a. $(-4099)^{1/5}$

b. $(-10,596)^{1/8}$

c. $(-5403)^{1/6}$

d. $(-46)^{1/2}$

Explanation:

Step1: Recall root properties

For a real - valued \(n\)th root of a number \(a\), i.e., \(a^{1/n}=\sqrt[n]{a}\):

• If \(n\) is odd, then \(\sqrt[n]{a}\) is a real number for any real number \(a\) (because for an odd \(n\), we can have a negative number inside the \(n\)th root and still get a real result. For example, \(\sqrt[3]{-8}=- 2\) since \((-2)^3=-8\)).
• If \(n\) is even, then \(\sqrt[n]{a}\) is a real number only when \(a\geq0\) (because for an even \(n\), the \(n\)th power of a real number is non - negative. So, if \(a < 0\) and \(n\) is even, \(\sqrt[n]{a}\) is not a real number. For example, \(\sqrt{-4}\) is not a real number as there is no real number \(x\) such that \(x^{2}=-4\)).

Step2: Analyze each option

• Option A: For \((-4099)^{1/5}=\sqrt[5]{-4099}\). Here, \(n = 5\) (which is odd). By the property of \(n\)th roots, when \(n\) is odd, the \(n\)th root of a negative number is a real number. Let \(x=\sqrt[5]{-4099}\), then \(x^{5}=-4099\), and \(x =-\sqrt[5]{4099}\), which is a real number.
• Option B: For \((-10596)^{1/8}=\sqrt[8]{-10596}\). Here, \(n = 8\) (which is even) and the number inside the root \(a=-10596<0\). So, \(\sqrt[8]{-10596}\) is not a real number.
• Option C: For \((-5403)^{1/6}=\sqrt[6]{-5403}\). Here, \(n = 6\) (which is even) and \(a=-5403<0\). So, \(\sqrt[6]{-5403}\) is not a real number.
• Option D: For \((-46)^{1/2}=\sqrt{-46}\). Here, \(n = 2\) (which is even) and \(a = - 46<0\). So, \(\sqrt{-46}\) is not a real number.

Answer:

A. \((-4099)^{1/5}\)