There are TWO solutions for x, not just one.  To do this you
need to know the change of base formula 

and the definition of a logarithm:   where c is the exponent to
which base b must be raised to give a.  That is  is equivalent
to the equation   


log3x + logx9 = 3

Use the change of base formula on the second term:

log3x +  = 3

Multiply through by log3x

(log3x)² + log39 = 3·log3x

log39 = 2 since the exponent to which the base 3 must be
raised to give 9 is the exponent 2, since 3² = 9 

(log3x)² + 2 = 3·log3x

(log3x)² - 3·log3x + 2 = 0

That factors as a quadratic in log3x

(log3x - 2)(log3x - 1) = 0

Use the zero-factor property:

log3x - 2 = 0;   log3x - 1 = 0;
    log3x = 2;       log3x = 1;
        x = 32          x = 31
        x = 9           x = 3

Two solutions: 9 and 3.

Checking x = 9

log3x + logx9 = 3
log39 + log99 = 3
  2 + 1 = 3
      3 = 3

Checking x = 3

log3x + logx9 = 3
log33 + log39 = 3
  1 + 2 = 3
      3 = 3

Both answers check.

Edwin