Question 202721
This much of your solution is good:
{{{2/(2-log(3,x))=6/(3+log(3,x))-3}}}
assume: {{{log(3,x)=z}}} 
giving:  2/(2-z)=6/(3+z)-3<br>
Then we get a little off track finding the common denominator and subtracting. Since subtractions can cause so many problems I strongly encourage you to change subtractions to additions. So I would rewrite this as:
{{{2/(2 + (-z))=6/(3+z) + (-3)}}}
Now when we get the common denominator on the right side we might avoid the error you had:
{{{2/(2 + (-z))=6/(3+z) + (-3)*((3 + z)/(3 + z))}}}
{{{2/(2 + (-z))=6/(3+z) + ((-9) + (-3z))/(3 + z))}}}
{{{2/(2 + (-z))=((-3) + (-3z))/(3 + z))}}}
We can solve this by cross-multiplying:
{{{2*(3 + z) = (2 + (-z))*((-3) + (-3z))}}}
{{{6 + 2z = -6 + -6z + 3z + 3z^2}}}
{{{6 + 2z = 3z^2 + (-3z) + (-6)}}}
Subtracting 2z and 6 from both sides we get:
{{{0 = 3z^2 + (-5z) + (-12)}}}
This will factor:
{{{0 = (z - 3)(3z + 4)}}}
This gives solutions of z = 3 or z = -4/3.
Substituting back in for z we get:
{{{log(3,x) = 3}}} or {{{log(3,x) = (-4)/3}}}
Rewriting these in exponential form we get
{{{x = 3^3}}} or {{{x = 3^(-4/3)}}}
{{{x = 27}}} or {{{x = 1/(3^(4/3))}}}
{{{x = 27}}} or {{{x = 1/(root(3, 3^4))}}}
If we want rationalized denominators we need a perfect cube in the denominator of the second solution:
{{{x = 27}}} or {{{x = (1/(root(3, 3^4)))*((root(3, 3^2))/(root(3, 3^2)))}}}
{{{x = 27}}} or {{{x = (root(3, 3^2))/(root(3, 3^6))}}}
{{{x = 27}}} or {{{x = (root(3, 3^2))/(3^2)}}}
{{{x = 27}}} or {{{x = (root(3, 9))/9 = (1/9)root(3, 9)}}}<br>
Both of these answers check. (One should always check because we need to avoid extraneous solutions like those that would make the argument of a log function negative, make the radicand of an even-numbered root negative, denominators zero, etc.)