Four different interpretations of the question and four possible solutions:
- Assume (3^(x-3))^1/2=3^(1-6x); squaring both sides: 3^(x-3)=3^2(1-6x); equating exponents: x-3=2-12x, 13x=5, x=5/13.
- Assume 3^(x-3)^1/2=3^(1-6x); (x-3)^1/2=1-6x; x-3=(1-6x)^2=1-12x+36x^2; 36x^2-12x+1=x-3; 36x^2-13x+4=0 has only complex roots.
- Assume (3x-3)^1/2=3^(1-6x); 3x-3=3^(2-12x); x-1=3^(2-12x-1)=3^(1-12x); x=1+3^(1-12x); approximate solution: x=1 because 3^-11=1/177147=0.000005645, which is very small.
- Assume (3x-3)^1/2=(3^1)-6x=3-6x; 3(x-1)=(3(1-2x))^2=9(1-4x+4x^2); x-1=3(1-4x+4x^2); x-1=3-12x+12x^2; 12x^2-13x+4=0 has only complex roots.