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log3x - 2logx3 = 1
so
1/logx3 - 2logx3 = 1
Let y = logx3
then 1/y - 2y = 1
multiply through by y: 1 - 2y2 = y
that is, 2y2 + y - 1 = 0
which factorises to (y + 1)(2y -1) = 0
so that y = -1 or y = 1/2
y = -1 implies logx3 = -1 so that x-1 = 3 ie 1/x = 3 or x= 1/3
y = 1/2 implies logx3 = 1/2 so that x1/2 = 3 or x = 32 = 9
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