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

`log_ a b = 1/(log_b a)`

Reasoning by analogy yields:

`log_x 3 = 1/(log_3 x)`

`log_(sqrt x) 3 = 1/(log_3 sqrt x) => log_(sqrt x) 3 = 1/(log_3 (x^(1/2)))`

Using the power property of logarithms, yields:

`log_(sqrt x) 3 = 1/((1/2)log_3 x)`

You need to re-write the equation, such that:

`1/(log_3 x) + 2/(log_3 x) = 15`

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

`log_3 x = t`

Changing the variable in equation, yields:

`1/t + 2/t = 15 => 1 + 2 = 15t => 3 = 15t => t = 3/15 => t = 1/5`

Replacing back the original variable, yields:

`log_3 x = 1/5 => x = 3^(1/5) => x = root(5) 3`

Since the value `x = root(5) 3` is positive, you do not need to test its validity.

**Hence, evaluating the solution to the given equation, yields **`x = root(5) 3.`