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prealgebra intermediate

Problem

The product of

34

and

48

equals

(A)

712

(B)

2712

(C)

732

(D)

1232

(E)

21232

Solution

34

and

48

can be expressed as

2^{\frac{2}{3}}

and

2^{\frac{3}{4}}

, so their product is

2^{\frac{17}{12}} = 21232

.

Final answer

E