Posted by sarah on Monday, October 18, 2010 at 10:57pm.
log3 [ 2x(2x-3) ] = log3[ 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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