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 Algebra Pre-Algebra and Basic Algebra Math Forum

 April 30th, 2014, 01:42 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 solving log(21+23x+6x^2)/log(3+2x)=4-log(4x^2+12x+9)/log(7+3x) Last edited by skipjack; April 30th, 2014 at 12:16 PM.
 April 30th, 2014, 05:32 AM #2 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 x>-1 there's no solution.
 April 30th, 2014, 12:08 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,820 Thanks: 2159 Really? Then what happens for x = -1/4?
April 30th, 2014, 07:01 PM   #4
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Joined: Sep 2013
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Quote:
 Originally Posted by skipjack Really? Then what happens for x = -1/4?
Can you show me the solution? Thanks
My level is not that high. Teach me. thanks.

 April 30th, 2014, 07:43 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond $\displaystyle \frac{\log(6x^2+23x+21)}{\log(2x+3)}=4-\frac{\log(4x^2+12x+9)}{\log(3x+7)}$ $\displaystyle \frac{\log(3x+7)+\log(2x+3)}{\log(2x+3)}=4-\frac{2\log(2x+3)}{\log(3x+7)}$ $\displaystyle \frac{\log(3x+7)}{\log(2x+3)}=3-\frac{2\log(2x+3)}{\log(3x+7)}$ $\displaystyle u=\frac{\log(3x+7)}{\log(2x+3)}$ $\displaystyle u+2u^{-1}=3$ $\displaystyle u^2+2=3u$ $\displaystyle u^2-3u+2=0$ $\displaystyle (u-1)(u-2)=0$ $\displaystyle \frac{\log(3x+7)}{\log(2x+3)}=1\Rightarrow x=-4$ ...no solution. $\displaystyle \frac{\log(3x+7)}{\log(2x+3)}=2\Rightarrow x=-2,\,-\frac14$ ...x = -1/4 is a solution.
 April 30th, 2014, 08:28 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra No need for two copies of that.

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