When you evaluate any `log_a b` where a > b, a `!=` 1, then a) `log_a b` > 1 b) 0 < `log_a b` < 1 c) `log _ ab` < 0 d) Cannot be determined
When `a!=1` and a > b, `log_a b` can take on any value for different values of a and b.
`log_a b` can take on all values including 0 as there is no restriction in place that does not allow b to take on the value 1.
For a = 12, b = 10, `log_12 10 ~~ 0.9266`
For a = 0.8, b = 0.6, `log_0.8 0.6 ~~ 2.2892`
For a = 8, b = 0.1, `log_ 8 0.1 ~~ -1.107`
For a = 8, b = 1, `log_8 1 = 0`
The correct option is D.
`a>b``` so `ln(a)>ln(b)`
`=> x in(-oo,1)`
`=> log_a b in(-oo,1)`
Thus option A is not possible
Options B and C are correct, so we can not decide either B or C.
Thus correct answer for this multiple choice question will be D.