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

log3 2 +log3 X +log3 (2x-3)=log3 X + log3 (x+1) + log3 (x-1) - log3 (x^2-1)

- precalc -
**Reiny**, Monday, October 18, 2010 at 11:33pm
log_{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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