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)=4log(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 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36  
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 
$\displaystyle \frac{\log(6x^2+23x+21)}{\log(2x+3)}=4\frac{\log(4x^2+12x+9)}{\log(3x+7)}$ $\displaystyle \frac{\log(3x+7)+\log(2x+3)}{\log(2x+3)}=4\frac{2\log(2x+3)}{\log(3x+7)}$ $\displaystyle \frac{\log(3x+7)}{\log(2x+3)}=3\frac{2\log(2x+3)}{\log(3x+7)}$ $\displaystyle u=\frac{\log(3x+7)}{\log(2x+3)}$ $\displaystyle u+2u^{1}=3$ $\displaystyle u^2+2=3u$ $\displaystyle u^23u+2=0$ $\displaystyle (u1)(u2)=0$ $\displaystyle \frac{\log(3x+7)}{\log(2x+3)}=1\Rightarrow x=4$ ...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.
