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If 64a = \(\frac{1}{256^b}\), then 3a+4b equals

(a) 2 

(b) 4 

(c) 8 

(d) 0

1 Answer

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Best answer

(d) 0

 64a = \(\frac{1}{256^b}\) ⇒ (26)\(\frac{1}{(2^8)^b}\)

⇒ 26a × 28b = 1 ⇒ 26a + 8b = 20

⇒ 6a + 8b = 0 ⇒ 3a + 4b = 0

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