precalc
posted by sarah .
log3 2 +log3 X +log3 (2x3)=log3 X + log3 (x+1) + log3 (x1)  log3 (x^21)

log_{3} [ 2x(2x3) ] = log_{3}[ x(x+1)(x1)/(x^2  1) ]
"antilog" it , remember the old "whatever you do to one side, you must do to the other side"
2x(2x3) = x(x+1)(x1)/(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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