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cubicle_bound_misfit: Master | Next Rank: 500 Posts; Posts: 246; Joined: Mon May 19, 2008 7:34 am; Location: Texaco Gas Station; Thanked: 7 times

by cubicle_bound_misfit » Tue Oct 07, 2008 11:58 am

question is x^y < y^x

1) x = y^2

question becomes

y ^2y < y ^ (y^2)

taking log froom both side

is 2y < y^2

for y <0 and y> 2 it is true

but not for y =1 or 2

stmt 2

y> 2

so it would be true if we combine both statements together.

Hence C.

Cubicle Bound Misfit

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