If $3^{2x+1}-3^x=3^{x+3}-3^2$, then the values of $x$ are

1. Rewrite the exponential terms using the laws of exponents:
$$3\cdot (3^x{)}^2-3^x=27\cdot 3^x-9$$
Let $3^x=y$ (where $y>0$). Substitute $y$ into the equation to form a quadratic equation:
$$3y^2-y=27y-9$$
$$3y^2-28y+9=0$$
Solve the quadratic equation by factoring:
$$3y^2-27y-y+9=0$$
$$3y(y-9)-1(y-9)=0$$
$$(3y-1)(y-9)=0$$
This gives $y=9$ or $y=\frac{1}{3}$.Substitute $3^x=y$ back into the solutions to solve for $x$:Case 1: $3^x=9⟹3^x=3^2⟹x=2$ Case 2: $3^x=\frac{1}{3}⟹3^x=3^{-1}⟹x=-1$ Thus, the values of $x$ are $2,-1$.