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 
Math Team Joined: Jul 2011 From: Texas Posts: 2,750 Thanks: 1400  
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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