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 December 9th, 2008, 09:40 PM #1 Senior Member   Joined: Sep 2008 Posts: 199 Thanks: 0 exponents If $a^{2x-1}=b^{1-3y}$and $a^{3x-1}=b^{2y-2}$ , prove that $13xy-7x-5y+3=0$
December 10th, 2008, 01:10 AM   #2
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Re: exponents

Quote:
 Originally Posted by mikeportnoy If $a^{2x-1}=b^{1-3y}$and $a^{3x-1}=b^{2y-2}$ , prove that $13xy-7x-5y+3=0$
$\frac{(2x-1)ln(a)}{(3x-1)ln(a)}=\frac{(1-3y)ln(b)}{(2y-2)ln(b)}$

$(2x-1)(2y-2)=(1-3y)(3x-1)$

$4xy-4x-2y+2=3x-1-9xy+3y$

$13xy-7x-5y+3=0$

This completes the proof.

 December 10th, 2008, 07:14 AM #3 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: exponents Logs are not necessary. Just use rules for exponents.

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