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Christophicus
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#1
Prove that if a^x = b^y = (ab)^xy then x + y = 1
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Gauss
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(Original post by Widowmaker)
Prove that if a^x = b^y = (ab)^xy then x + y = 1
(ab)^xy

= [(ab)^x]^y
= (a^x . b^x)^y
= (b^y . b^x)^y
= [b^(x+y)]^y
= b^y(y+x)

But we know that b^y = (ab)^xy

=> b^y(y + x) = b^y
=> y = y(y + x)
=> 0 = y(y + x -1)
=> y + x -1 = 0
=> y + x = 1 QED

Euclid.
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