9^(2x)*27^(x^2)=3^(-1) solve the exponential equation

Question

The answer is -1/3,-1 but i dont know how to get it

Answer 1

9=3^2 and 27=3^3. So we rewrite the equation:

3^(4x)*3^(3x^2)=3^(-1); therefore 4x+3x^2=-1, equating exponents.

3x^2+4x+1=0=(3x+1)(x+1)=0, so x=-1/3 or -1.

CHECK: x=-1: 9^(-2)*27=27/81=1/3=3^(-1). OK.

9^(-2/3)*27^(1/9)=1/cuberoot(81)*(9√)(27); 3^(-4/3)*3^(1/3)=3^(-4/3+1/3)=3^(-1). OK