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April 24th, 2015, 02:55 AM   #1
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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

...
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April 24th, 2015, 04:09 AM   #2
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anyone!?
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April 24th, 2015, 05:44 AM   #3
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Quote:
Originally Posted by euphmorning View Post
anyone!?
Don't bump ... it's considered rude.

$2xy^2=32$

$2(xy)y=32$

since $xy=32$ ...

64y=32

can you finish?
Thanks from topsquark
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April 28th, 2015, 06:31 AM   #4
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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.
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April 28th, 2015, 06:49 AM   #5
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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.
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April 29th, 2015, 02:13 AM   #6
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Thanks, silly error on my part. got it fine now.
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