A)

log_{7} 10 = 1 + log_{7} (10/7) = 1 + log_{7} 1.428

log_{11} 13 = 1 + log_{11} (13/11) = 1 + log_{11 } 1.18

Let A = log_{7} 1.428

and B = log_{11} 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 + log_{7} 1.428 > 1 + log_{11} 1.18

________________________________________________________

## No comments:

Post a Comment