1. **State the problem:** Solve the equation $\log_{x+8}(3x^2 + 8x) = 2$ for $x$. 2. **Recall the logarithm definition:** $\log_a b = c$ means $a^c = b$, where $a > 0$, $a \neq 1$, and $b > 0$. 3. **Apply the definition:** From $\log_{x+8}(3x^2 + 8x) = 2$, we get $$ (x+8)^2 = 3x^2 + 8x $$ 4. **Expand the left side:** $$ (x+8)^2 = x^2 + 16x + 64 $$ 5. **Set up the equation:** $$ x^2 + 16x + 64 = 3x^2 + 8x $$ 6. **Bring all terms to one side:** $$ x^2 + 16x + 64 - 3x^2 - 8x = 0 $$ $$ -2x^2 + 8x + 64 = 0 $$ 7. **Multiply both sides by $-1$ to simplify:** $$ \cancel{-1} \times (-2x^2 + 8x + 64) = \cancel{-1} \times 0 $$ $$ 2x^2 - 8x - 64 = 0 $$ 8. **Divide entire equation by 2:** $$ \cancel{2}x^2 - \cancel{2} \times 4x - \cancel{2} \times 32 = 0 $$ $$ x^2 - 4x - 32 = 0 $$ 9. **Solve quadratic equation using the quadratic formula:** $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=1$, $b=-4$, $c=-32$. 10. **Calculate discriminant:** $$ \Delta = (-4)^2 - 4 \times 1 \times (-32) = 16 + 128 = 144 $$ 11. **Find roots:** $$ x = \frac{4 \pm \sqrt{144}}{2} = \frac{4 \pm 12}{2} $$ 12. **Calculate each root:** - $x = \frac{4 + 12}{2} = \frac{16}{2} = 8$ - $x = \frac{4 - 12}{2} = \frac{-8}{2} = -4$ 13. **Check domain restrictions:** - Base $x+8 > 0 \Rightarrow x > -8$ - Base $x+8 \neq 1 \Rightarrow x \neq -7$ - Argument $3x^2 + 8x > 0$ 14. **Check argument for $x=8$:** $$3(8)^2 + 8(8) = 3 \times 64 + 64 = 192 + 64 = 256 > 0$$ valid. 15. **Check argument for $x=-4$:** $$3(-4)^2 + 8(-4) = 3 \times 16 - 32 = 48 - 32 = 16 > 0$$ valid. 16. **Check base for $x=8$:** $$8 + 8 = 16 > 0$$ and $16 \neq 1$ valid. 17. **Check base for $x=-4$:** $$-4 + 8 = 4 > 0$$ and $4 \neq 1$ valid. 18. **Both solutions satisfy domain restrictions.** **Final answer:** $$ x = 8 \text{ or } x = -4 $$