another log question

[Bryn]

how would you go about solving log_3x - 2log_x3 = 1

[e(ho0n3]

Make the substitution
\log_x{3} = \log_3{3}/\log_3{x}
Manipulate the equation until you get
(\log_3{x})^2 - \log_3{x} - 2 = 0
Then make the substitution
u = \log_3{x}
Solve for u, then solve for x.