Algebra Pre-Algebra and Basic Algebra Math Forum

 April 30th, 2014, 02:42 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 01:16 PM. April 30th, 2014, 06:32 AM #2 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 x>-1 there's no solution. April 30th, 2014, 01:08 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 Really? Then what happens for x = -1/4? April 30th, 2014, 08:01 PM   #4
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 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, 08:43 PM #5 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,974 Thanks: 1156 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, 09:28 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra No need for two copies of that. Tags soling, solving Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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