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 April 24th, 2015, 01:55 AM #1 Newbie   Joined: Jul 2012 Posts: 18 Thanks: 0 exponents Hi there! ok so given: x=2^p and y=2^q how on earth do i solve for p and q for the equations: xy=32 and 2x(y^2)=32 this has befuddled me completely. help appreciated!!! many thanks i know that they are equivalent.. so commencing things we have: xy=2x(y^2)=32 ...
 April 24th, 2015, 03:09 AM #2 Newbie   Joined: Jul 2012 Posts: 18 Thanks: 0 anyone!?
April 24th, 2015, 04:44 AM   #3
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Quote:
 Originally Posted by euphmorning anyone!?
Don't bump ... it's considered rude.

$2xy^2=32$

$2(xy)y=32$

since $xy=32$ ...

64y=32

can you finish?

 April 28th, 2015, 05:31 AM #4 Newbie   Joined: Jul 2012 Posts: 18 Thanks: 0 My apologies, I didn't intend to be rude :/ OK so I think I have it.. q=-1 p= 4 Have I got this correct? Thankyou again.
 April 28th, 2015, 05:49 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,159 Thanks: 866 No, it is not! That easy to check. If q= -1 and p= 4 then x= 2^4= 16 and y= 2^q= 1/2. xy= 16/2= 8 not 32. Another way of looking at it: With x= 2^p and y= 2^q, xy= 2^p2^q= 2^(p+q)= 32= 2^5 so p+ q= 5. Also 2xy^2= 2^12^p2^(2q)= 2^(1+ p+ 2q)= 32= 2^5 so 1+ p+ 2q= 5 and p+ 2q= 4. Solve the equations p+ q= 5, p+ 2q= 4.
 April 29th, 2015, 01:13 AM #6 Newbie   Joined: Jul 2012 Posts: 18 Thanks: 0 Thanks, silly error on my part. got it fine now.

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