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Saturday, December 13, 2008

Q) Prove that .

A)

log7 10 = 1 + log7 (10/7) = 1 + log7 1.428

log11 13 = 1 + log11 (13/11) = 1 + log11 1.18


Let A = log7 1.428


and B = log11 1.18


So, 7A = 1.428


and 11B = 1.18


Clearly A and B will both be fractional, and B will be smaller than A; (Since, in case of A, 7 is being reduced to 1.428 but with B, a larger number 11 is being reduced to a correspondingly less value than for A. )


Thus, 1 + log7 1.428 > 1 + log11 1.18

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