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prealgebra intermediate

Problem

If

N > 1

, then

3 N 3 N 3 N =

(A)

N^{\frac{1}{27}}

(B)

N^{\frac{1}{9}}

(C)

N^{\frac{1}{3}}

(D)

N^{\frac{13}{27}}

Solution

The key identities are

3 x^{n} = x^{\frac{n}{3}}

and

x \cdot x^{\frac{a}{b}} = x^{1 + \frac{a}{b}}

3 N 3 N 3 N

3 N 3 N \cdot N^{\frac{1}{3}}

3 N 3 N^{\frac{4}{3}}

3 N \cdot N^{\frac{4}{9}}

3 N^{\frac{13}{9}}

N^{\frac{13}{27}}

, thus the answer is

N^{\frac{13}{27}}

Final answer

D