Question

1. mrs. tarpley wrote the expression ((27x)^{\frac{1}{3}}) on the board. three students made statements about the expression. which student made a correct statement?

dorie
the expression is equivalent to (\frac{1}{(27x)^3}).
alonso
the expression is equivalent to (9(sqrt3{x})).
nick
the expression is equivalent to the cube root of (27x).

Explanation:

Step1: Recall exponent rules

We know that \(a^{\frac{1}{n}}=\sqrt[n]{a}\) and \((ab)^n = a^n b^n\), also \(a^{-n}=\frac{1}{a^n}\).

Step2: Analyze the original expression \((27x)^{\frac{1}{3}}\)

First, use the power - of - a - product rule \((ab)^n=a^n\times b^n\). So \((27x)^{\frac{1}{3}}=27^{\frac{1}{3}}\times x^{\frac{1}{3}}\).

Since \(27 = 3^3\), then \(27^{\frac{1}{3}}=(3^3)^{\frac{1}{3}}\). According to the power - of - a - power rule \((a^m)^n=a^{m\times n}\), we have \((3^3)^{\frac{1}{3}} = 3^{3\times\frac{1}{3}}=3\).

And \(x^{\frac{1}{3}}=\sqrt[3]{x}\), so \((27x)^{\frac{1}{3}}=3\times\sqrt[3]{x}=3\sqrt[3]{x}\).

Now let's analyze each student's statement:

• Dorie's statement: \(\frac{1}{(27x)^3}=(27x)^{-3}\), which is not equal to \((27x)^{\frac{1}{3}}\) because the exponents \(- 3\) and \(\frac{1}{3}\) are different.
• Alonso's statement: We just found that \((27x)^{\frac{1}{3}} = 3\sqrt[3]{x}\), so Alonso's statement (assuming the typo "d" is a mistake and it should be "3") is correct.
• Nick's statement: The cube root of \(27x\) is \(\sqrt[3]{27x}\), and \(\sqrt[3]{27x}=\sqrt[3]{27}\times\sqrt[3]{x}=3\sqrt[3]{x}=(27x)^{\frac{1}{3}}\), but Nick's statement description is a bit unclear. However, from the calculation, Alonso's statement (after correcting the possible typo) is in line with the equivalent form of \((27x)^{\frac{1}{3}}\).

Answer:

Alonso (assuming the "d" in Alonso's statement is a typo and should be "3") made a correct statement. If we consider the options as per the given (with possible typos), the student with the correct statement is Alonso.