The equation `2 - log_3 2 = log_3 x` has to be solved for x.

`2 - log_3 2 = log_3 x`

Take the logarithmic terms to one side

`2 = log_3 2 + log_3 x`

Now use the property log a + log b = log(a*b)

`2 = log_3(2*x)`

If log_b x = y, x = b^y

3^2 = 2*x

9 = 2x

x = 9/2

The solution of the given equation is x = 9/2 = 4.5

We have to solve for x: 2 - log 3 2 = log 3 x

Now we use the relation for algorithms that log (a*b) = log a + log b.

2 - log 3 2 = log 3 x

=> 2 = log 3 x + log 3 2

=> 2 = log 3 x*2

now 2 = log 3 3^2 , as log a^b = b log a and log 3 (3) = 1.

=> 3^2 = 2*x

=> x = 3^2 / 2

=> x = 9/2

**Therefore x = 9/2**.

For the logarithm to exist, x has to be positive.

We'll write 2 as:

2*1 = 2* log 3 3

We'll use the power property of logarithms and the symmetric property:

log 3 (x) = log 3 (3)^2 - log 3 (2)

Because the bases are matching, we'll transform the difference of logarithms from the right side, into a quotient. We'll apply the formula:

lg a - lg b = lg (a/b)

We'll substitute a by 9 and b by 2. The logarithms from formula are decimal logarithms. We notice that the base of logarithm is 3.

log 3 (x) = log 3 (9/2)

Because the bases are matching, we'll apply the one to one property:

x = 9/2

**x = 4.5**

**Since the value of x is positive, the solution of the equation is admissible.**