What is x when log base 3 x - 2log base x 3 equals 1?

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