Question 135665
1n(3x-1)=1n 1 -1n (x-1) 
ln(3x-1) = 0 -ln(x-1)
ln[(3x-1)(x-1)] = 0
3x^2-4x+1 = 1
3x^2-4x = 0
x(3x-4)=0
x = 0 or  4/3
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But x cannot be 0 for that would give us ln(0-1) = ln(-1), which is meaningless.
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So the only solution is x = 4/3
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Cheers,
Stan H.