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Re: If xaybzc equals the product of 154 and 56, z > y > x, and a
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12 Nov 2016, 16:40
Explanation
To solve this question, find the prime factors: The prime factors of 154 are 2, 7, and 11, and the prime factors of 56 are 2, 2, 2, and 7. Thus, the product of 154 and 56 will have the prime factors 2, 2, 2, 2, 7, 7, and 11, or (\(2^4\))(\(7^2\))(\(11^1\)).
Line up your bases and exponents with the inequalities, and you get a = 4, b = 2, and c = 1 for the bases, and x = 2, y = 7, z = 11 for the exponents.
Now \(a^xb^yc^z\) = \((4^2)(2^7)(1^1^1)\), which equals 16 × 128 × 1, or 2,048. The correct answer is choice B.