When you evaluate any `log_(a)b`
where a > b and b > 1 we use the definition of logs, which is:
`log_(a)b = x ` is `a^(x) = b`
If `x < 0;`
then b will be less than 1, however our restrictions state that b > 1.
For instance, if `5^-1 = .2`
If x = 0, then b = 1, therefore our answer won't include 0. Also, using the same example: `5^1 = 5,`
therefore x must have a value less than 1 as 1 makes a = b, but we want a> b. For instance `5^0.9 ~~4.257`
Therefore, our correct answer would be choice "b".