Question

which exponential expression is equivalent to
sqrt8{y^3}?

options:
(y^{\frac{3}{8}})
(y^{\frac{8}{3}})
(\frac{1}{y^{\frac{8}{3}}})
(\frac{1}{y^{\frac{3}{8}}})

Explanation:

Step1: Recall radical to exponent rule

The \(n\)-th root of a number \(a\) can be written as \(a^{\frac{1}{n}}\), and \(\sqrt[m]{a^n}=a^{\frac{n}{m}}\). For a nested radical \(\sqrt[8]{\sqrt{y^3}}\), first handle the inner square root. The square root of \(y^3\) is \(y^{\frac{3}{2}}\) (since \(\sqrt{y^3}=y^{\frac{3}{2}}\) by \(\sqrt[m]{a^n}=a^{\frac{n}{m}}\) with \(m = 2\), \(n=3\)).

Step2: Handle the outer 8 - th root

Now we have \(\sqrt[8]{y^{\frac{3}{2}}}\). Using the rule \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\) again, where \(n = 8\) and \(m=\frac{3}{2}\). So we calculate \(\frac{\frac{3}{2}}{8}=\frac{3}{2}\times\frac{1}{8}=\frac{3}{16}\)? Wait, no, wait. Wait, the original is \(\sqrt[8]{\sqrt{y^3}}\), which is \((y^3)^{\frac{1}{2}\times\frac{1}{8}}\)? Wait, no, let's re - express step by step.

First, \(\sqrt{y^3}=(y^3)^{\frac{1}{2}}\) (by the definition of square root: \(\sqrt{a}=a^{\frac{1}{2}}\)). Then \(\sqrt[8]{\sqrt{y^3}}=\sqrt[8]{(y^3)^{\frac{1}{2}}}\). Now, using the property of exponents \((a^m)^n=a^{m\times n}\) and \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\), we have \((y^3)^{\frac{1}{2}\times\frac{1}{8}}=y^{3\times\frac{1}{2}\times\frac{1}{8}}\)? No, wait, \(\sqrt[8]{(y^3)^{\frac{1}{2}}}=(y^3)^{\frac{1}{2}\times\frac{1}{8}}\)? No, the rule for \(\sqrt[n]{a^m}\) is \(a^{\frac{m}{n}}\), so if \(a=(y^3)^{\frac{1}{2}}\) and \(n = 8\), then \(\sqrt[8]{(y^3)^{\frac{1}{2}}}=((y^3)^{\frac{1}{2}})^{\frac{1}{8}}\). Then by the power of a power rule \((a^m)^n=a^{m\times n}\), we get \(y^{3\times\frac{1}{2}\times\frac{1}{8}}\)? Wait, no, that's wrong. Wait, \(\sqrt{y^3}=y^{\frac{3}{2}}\) (because \(\sqrt{a}=a^{\frac{1}{2}}\), so \(\sqrt{y^3}=(y^3)^{\frac{1}{2}}=y^{3\times\frac{1}{2}}=y^{\frac{3}{2}}\)). Then \(\sqrt[8]{y^{\frac{3}{2}}}\) is equal to \(y^{\frac{3}{2}\times\frac{1}{8}}\)? No, \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\), so here \(n = 8\) and \(m=\frac{3}{2}\), so \(\sqrt[8]{y^{\frac{3}{2}}}=y^{\frac{\frac{3}{2}}{8}}=y^{\frac{3}{2}\times\frac{1}{8}}=y^{\frac{3}{16}}\)? Wait, that's not matching the options. Wait, maybe I misread the original problem. Wait, the original problem is \(\sqrt[8]{\sqrt{y^3}}\) or is it \(\sqrt[8]{y^3}\)? Wait, the user wrote " \(\boldsymbol{\sqrt[8]{\sqrt{y^3}}}\) "? Wait, no, looking at the image description: "Which exponential expression is equivalent to \(\boldsymbol{\sqrt[8]{\sqrt{y^3}}}\)?" Wait, no, maybe it's \(\sqrt[8]{y^3}\)? Wait, the options have \(y^{\frac{3}{8}}\), \(y^{\frac{8}{3}}\), \(\frac{1}{y^{\frac{8}{3}}}\), \(\frac{1}{y^{\frac{3}{8}}}\). Wait, if the problem is \(\sqrt[8]{y^3}\), then by the rule \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\), so \(\sqrt[8]{y^3}=y^{\frac{3}{8}}\). Let's check: \(\sqrt[8]{y^3}=(y^3)^{\frac{1}{8}}=y^{3\times\frac{1}{8}}=y^{\frac{3}{8}}\), which is the first option.

Wait, maybe the original problem was \(\sqrt[8]{y^3}\) (maybe a typo in the square root of square root). Let's proceed with that.

Step1: Recall the radical - exponent conversion rule

The \(n\) - th root of \(a^m\) is given by \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\), where \(a\) is a real number, \(m\) and \(n\) are positive integers, and \(n>1\).

Step2: Apply the rule to \(\sqrt[8]{y^3}\)

For the expression \(\sqrt[8]{y^3}\), we have \(a = y\), \(m = 3\), and \(n = 8\). Using the rule \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\), we substitute the values:
\(\sqrt[8]{y^3}=y^{\frac{3}{8}}\)

Answer:

\(y^{\frac{3}{8}}\) (the first option, green box)