You need to move all terms to one side, such that:

`(log_3 x)^2 - log_3 x^2 - 3 = 0`

You need to come up with the following substitution, such that:

`log_3 x = t`

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

You may use quadratic formula to find the solutions to quadratic equation, such that:

`t_(1,2) = (-(-2)+-sqrt((-2)^2 - 4*1*(-3)))/(2*1)`

`t_(1,2) = (2 +- sqrt(4 + 12))/2`

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

You need to substitute back` log_3 x = t_(1,2)` such that:

`log_3 x = 3 => log_3 x = 3*log_3 3 => log_3 x = log_3 3^3 => x = 3^3 => x = 27`

`log_3 x = -1 => log_3 x = -1*log_3 3 => log_3 x = lo_3 3^(-1) => log_3 x = log_3 (1/3) => x = 1/3`

**Hence, evaluating the solutions to the given logarithmic equation yields `x = 1/3` and `x = 27` .**

Since we have the term log3 x^2, we'll use the power property of logarithms:log3 x^2 = 2 log3 x.

The given equation (log3 x)^2 = log3 x^2 + 3 converts to

(log3 x)^2 - 2log3 x - 3 = 0

We'll substitute log3 x = u.

We'll re-write the equation:

u^2 - 2u -3 =0

Factoring, we'll get:

(u-3)(u+1) =0

We'll put each factor as zero:

u-3 = 0

So, u = 3

u + 1 = 0

u = -1

But log3 x = 3 => x = 3^3 => x = 27

log3 x = -1 => x = 3^-1 => x = 1/3

Since both solutions are positive, we'll accept them.