# precalc

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

• precalc -

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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