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April 30th, 2014, 01:42 AM   #1
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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.
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April 30th, 2014, 05:32 AM   #2
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x>-1

there's no solution.
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April 30th, 2014, 12:08 PM   #3
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Really? Then what happens for x = -1/4?
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April 30th, 2014, 07:01 PM   #4
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Really? Then what happens for x = -1/4?
Can you show me the solution? Thanks
My level is not that high. Teach me. thanks.
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April 30th, 2014, 07:43 PM   #5
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$\displaystyle \frac{\log(6x^2+23x+21)}{\log(2x+3)}=4-\frac{\log(4x^2+12x+9)}{\log(3x+7)}$

$\displaystyle \frac{\log(3x+7)+\log(2x+3)}{\log(2x+3)}=4-\frac{2\log(2x+3)}{\log(3x+7)}$

$\displaystyle \frac{\log(3x+7)}{\log(2x+3)}=3-\frac{2\log(2x+3)}{\log(3x+7)}$

$\displaystyle u=\frac{\log(3x+7)}{\log(2x+3)}$

$\displaystyle u+2u^{-1}=3$

$\displaystyle u^2+2=3u$

$\displaystyle u^2-3u+2=0$

$\displaystyle (u-1)(u-2)=0$

$\displaystyle \frac{\log(3x+7)}{\log(2x+3)}=1\Rightarrow x=-4$ ...no solution.

$\displaystyle \frac{\log(3x+7)}{\log(2x+3)}=2\Rightarrow x=-2,\,-\frac14$ ...x = -1/4 is a solution.
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April 30th, 2014, 08:28 PM   #6
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No need for two copies of that.
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