Solution: We have ${{x}^{{{\log }_{3}}(x-1)}}=a\Rightarrow {{\log }_{3}}(x-1)\cdot {{\log }_{3}}x={{\log }_{3}}a$, and analogous ${{(x-1)}^{{{\log }_{3}}x}}=b\Rightarrow {{\log }_{3}}x\cdot {{\log }_{3}}(x-1)={{\log }_{3}}b$, thus we have $a=b$, se we get ${{x}^{{{\log }_{3}}(x-1)}}={{x}^{2}}\Rightarrow {{\log }_{3}}(x-1)=2\Rightarrow x=10$

My answer is also 10.I used DIFFERENT substitution .let log denote log base 3.Then let a=log(x).log(x-1); We get that the left hand side of the equation is 3^a+2.3^a=3.3^(2.(log x));From here we get after cancelling a 3 and comparing exponents that log(x).log(x-1)=2.(log x) From here we get that x-1=3^2 or x=10 EDIT: A perfect power post