- #31
mathwonk
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how's this?
A = 1/5 - 1/7 > 1/6 - 1/8 = a, hence (13)^(1/5) - (13)^(1/7)
=[(13)^(1/7][13^A - 1] > [(13)^(1/8)][13^a - 1]
= (13)^(1/6) - (13)^(1/8), hence
y = (13)^(1/5) + (13)^(1/8) > (13)^(1/6) + (13)^(1/7) = x.
just elementary precalculus, unless I made a mistake, which I frequently do in this realm.
edit: I guess this is basically the same as pasmith's (more general) solution in the previous post.
A = 1/5 - 1/7 > 1/6 - 1/8 = a, hence (13)^(1/5) - (13)^(1/7)
=[(13)^(1/7][13^A - 1] > [(13)^(1/8)][13^a - 1]
= (13)^(1/6) - (13)^(1/8), hence
y = (13)^(1/5) + (13)^(1/8) > (13)^(1/6) + (13)^(1/7) = x.
just elementary precalculus, unless I made a mistake, which I frequently do in this realm.
edit: I guess this is basically the same as pasmith's (more general) solution in the previous post.
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