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If 2k > 111111.11111 , then the minimum integral value of k that satisfies the given condition is (given that log 2 = 0.3010 and log 3 = 0.4771 ) |
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Answer» If 2k > 111111.11111 , then the minimum integral value of k that satisfies the given condition is (given that log 2 = 0.3010 and log 3 = 0.4771 ) |
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