You need to use the followng logarithmic identity, such that:

`log_a b = 1/(log_b a)`

Reasoning by analogy yields:

`log_x 3 = 1/(log_3 x)`

Replacing `1/(log_3 x)` for `log_x 3` yields:

`log_3 x - 1/(log_3 x) = 3/2`

You should come up with the following substitution, such that:

`log_3 x = t `

Replacing the variable, yields:

`t - 1/t = 3/2`

Bringing the terms to a common denominator, yields:

`2t^2 - 2 = 3t => 2t^2 - 3t - 2 = 0`

Using quadratic formula, yields:

`t_(1,2) = (3+-sqrt(9 + 16))/4`

`t_(1,2) = (3+-5)/4 => t_1 = 2 ; t_2 = -1/2`

You should replace back `log_3 x` for `t` , such that:

`log_3 x= 2 => x_1 = 3^2 => x_1 = 9`

`log_3 x= -1/2 => x_2 = 3^(-1/2) => x_2 = 1/(sqrt 3) => x_2 = sqrt3/3`

**Hence, evaluating the solutions to the given equation yields `x = sqrt3/3` and **`x = 9.`