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September 20, 2014

September 20, 2014

Posted by **sarah** on Monday, October 18, 2010 at 10:57pm.

- precalc -
**Reiny**, Monday, October 18, 2010 at 11:33pmlog

_{3}[ 2x(2x-3) ] = log_{3}[ x(x+1)(x-1)/(x^2 - 1) ]

"anti-log" it , remember the old "whatever you do to one side, you must do to the other side"

2x(2x-3) = x(x+1)(x-1)/(x^2 - 1)

4x^2 - 6x = x

4x^2 - 7x = 0

x(4x - 7) = 0

x = 0 or x = 7/4

but log x would be undefined if x=0

so x=7/4

better check my arithmetic

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