Suppose that a>b>1, then
[tex]a^x>b^x[/tex] and [tex]\log_ax<\log_bx.[/tex]
Therefore, since 2<3<7, [tex]\log_2x>\log_3x>\log_7x.[/tex]
Choose an arbitrary x>1. You have that a takes the greatest values at x, c takes the smallest value at x. Thus,
a>b>c and
- [tex]a=\log_2x;[/tex]
- [tex]b=\log_3x;[/tex]
- [tex]c=\log_7x.[/tex]
Answer: correct option is B.
