Solve for \( x \). \[ 27^{5 x+8}=3^{x^{2}+6 x+38} \] If there is more than one solution, separat

To solve the equation \[ 27^{5x + 8} = 3^{x^2 + 6x + 38}, \] we first express 27 as a power of 3: \[ 27 = 3^3. \] Then, we can rewrite \( 27^{5x + 8} \) as: \[ (3^3)^{5x + 8} = 3^{3(5x + 8)} = 3^{15x + 24}. \] Now we can rewrite the original equation: \[ 3^{15x + 24} = 3^{x^2 + 6x + 38}. \] Since the bases are the same, we equate the exponents: \[ 15x + 24 = x^2 + 6x + 38. \] Next, we rearrange the equation to set it to zero: \[ x^2 + 6x + 38 - 15x - 24 = 0, \] which simplifies to: \[ x^2 - 9x + 14 = 0. \] Now, we can factor the quadratic: \[ (x - 7)(x - 2) = 0. \] Setting each factor to zero gives us: \[ x - 7 = 0 \quad \Rightarrow \quad x = 7, \] \[ x - 2 = 0 \quad \Rightarrow \quad x = 2. \] Thus, the solutions to the equation are: \[ \boxed{7} \text{ and } \boxed{2}. \]