Before solving the equation, we'll impose constraints of existence of logarithms:

3+ log 3 x > 0 => log 3 x>-3

and

x>0

log 3 x>-3

log 3 x>-3*1

log 3 x>-3*log 3 3

log 3 x> log 3 3^-3

log 3 x> log 3 (1/27)

Because the base of logarithms is 3>1, the the function is increasing, so:

x > 1/27

The admissible solutions of x have to belong to the interval (1/27, inf.).

We'll solve the equation:

log 2 ( 3+ log 3 x) = 2

3+ log 3 x = 2^2

3+ log 3 x = 4

log 3 x = 4-3

log 3 x = 1

x = 3^1

**x = 3**

**Because 3 is in the interval (1/27 , inf.) is a valid solution.**

The solution of `log_2 ( 3+ log_3 x) = 2` has to be determined.

If `log_b x = y` , `x = b^y` .

Applying this to `log_2 ( 3+ log_3 x) = 2` gives:

`3+ log_3 x = 2^2`

`3+ log_3 x = 4`

`log_3 x = 4 -3`

`log_3 x = 1`

Again applying the relation given earlier, we get

x = 3^1

x = 3

The solution of the equation `log_2 ( 3+ log_3 x) = 2` is x = 3

log2 (3+log3x) =2.

To solve for x.

Solution:

log2 (3+ log3) = 2. Taking antologarithms with respect to base 2, we get:

(3+log3x ) = 2^2 = 4.

log3 x = 4-3 = 1. Take antilog (hope this is 10 base not 3 base).

3x = 10

x = 10/3.

If log3 x is a log to base e, then

log3 x = 1. Antilog wrt base 3 gives:

x = 3.