The least integer greater than each number for log₅1/47 = -3
All whole numbers and negative numbers are considered integers. This means that if we combine negative numbers with whole numbers, a collection of integers results.
Let y = log₅1/47
We know, logₐ x = y, only if a^y= x
Therefore, 5^y = 1/47
So, we must find an integer y such that 5 > 1/47
5^y > 1/47
5^{-3} > 1/47
(As, 5² = 25; 5³ = 125) (Since, 5² < 47, the answer will be 5³ as 5³ > 47)In this case, y = -3 will be the least integer to satisfy this inequality.
Therefore, the ultimate answer to the equation is log₅1/47 = -3
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