Hey BEN✌️, good to see you again. Step 1: Rewrite the right side of the equation with the same base as the left side. The equation is $3^{x^2+2} = 27^{2x-1}$. Since $27 = 3^3$, we can substitute this into the equation. $$3^{x^2+2} = (3^3)^{2x-1}$$ Step 2: Apply the exponent rule $(a^m)^n = a^{mn}$. $$3^{x^2+2} = 3^{3(2x-1)}$$ Step 3: Equate the exponents since the bases are the same. $$x^2+2 = 3(2x-1)$$ Step 4: Expand the right side of the equation. $$x^2+2 = 6x-3$$ Step 5: Rearrange the equation into a standard quadratic form $ax^2+bx+c=0$. $$x^2 - 6x + 2 + 3 = 0$$ $$x^2 - 6x + 5 = 0$$ Step 6: Solve the quadratic equation by factoring. We need two numbers that multiply to 5 and add to -6. These numbers are -1 and -5. $$(x-1)(x-5) = 0$$ Step 7: Set each factor equal to zero to find the values of $x$. $$x-1 = 0 \implies x = 1$$ $$x-5 = 0 \implies x = 5$$ The values of $x$ are $1$ and $5$. $$\boxed{x=1 \text{ or } x=5}$$ Send me the next one 📸