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, 7^A = 1.428

and 11^B = 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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