Monday, January 19, 2004

2004 #19 and 23

This year was 2004.  They had to work that in somehow.
Difficulty: easy.
Difficulty: Pretty complex. I was chasing log rules around for a while trying to get all this worked out.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest 2004

answer 19, answer 23.
Posted by at
Labels: ,

1 comment:

  1. PeggyUNovember 20, 2011 at 2:43 PM

    I think I might have found a shortcut on 23. Since y = x^(-1/4), then x^2 = y^-8; so logy(x^2/y^.5) = logy(y^(-8)/y^.5) = -17/2.

    Then you could have: logx (xy^5) - (-17/2) or logx(xy^5) + 17/2.

    Then logx x + 5logx y + 17/2 = 1 + 5(-1/4) + 17/2 = 33/4 :)

    ReplyDelete

Subscribe to: Post Comments (Atom)