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#940759 - 02/17/14 11:57 AM Be my EXponent!
CanukDenis Offline
Adept Boomer

Registered: 11/26/00
Posts: 14916
Loc: Ottawa Ontario Canada
Loved this one; looks complicated as old heck, but quite simple:

18^(6-3y) = 9^(2y-x-1)
Solve for x and y.
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#940764 - 02/17/14 01:00 PM Re: Be my EXponent! [Re: CanukDenis]
manxman Offline
Adept Boomer

Registered: 07/23/02
Posts: 13913
Loc: Unionville
Simplifying:
(2*9)^(6-3y)=9^(2y-x-1)
2^(6-3y)*9(6-3y)=9^(2y-x-1)
dividing both sides by 9^(6-3y)gives
2^(6-3y)=9^(5y-x-7)
since left side is even and right side is odd except when the exponents are zero then we solve 6-3y=0 and 5y-x-7=0 which gives x=3, y=2
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#940780 - 02/17/14 02:43 PM Re: Be my EXponent! [Re: CanukDenis]
CanukDenis Offline
Adept Boomer

Registered: 11/26/00
Posts: 14916
Loc: Ottawa Ontario Canada
Right!

Or 9^0 = 18^0
So 6-3y = 0 and 2y-x-1 = 0

3y = 6, so y = 2

4-x-1 = 0, so x = 3
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