1. **State the problem:** Solve the equation $$\log_{2x}(3x^2 + 10x - 16) = 2$$ for all values of $x$. 2. **Recall the logarithm definition:** If $$\log_a b = c$$, then $$a^c = b$$, where $a > 0$, $a \neq 1$, and $b > 0$. 3. **Apply the definition:** From $$\log_{2x}(3x^2 + 10x - 16) = 2$$, we get $$$(2x)^2 = 3x^2 + 10x - 16$$$ 4. **Simplify the equation:** $$$4x^2 = 3x^2 + 10x - 16$$$ 5. **Bring all terms to one side:** $$$4x^2 - 3x^2 - 10x + 16 = 0$$$ $$$x^2 - 10x + 16 = 0$$$ 6. **Factor the quadratic:** $$$x^2 - 10x + 16 = (x - 8)(x - 2) = 0$$$ 7. **Solve for $x$:** $$$x - 8 = 0 \Rightarrow x = 8$$$ $$$x - 2 = 0 \Rightarrow x = 2$$$ 8. **Check domain restrictions:** - Base of logarithm $2x$ must be positive and not equal to 1: $$$2x > 0 \Rightarrow x > 0$$$ $$$2x \neq 1 \Rightarrow x \neq \frac{1}{2}$$$ - Argument of logarithm $3x^2 + 10x - 16$ must be positive: $$$3x^2 + 10x - 16 > 0$$$ 9. **Check each solution:** - For $x=8$: $$$3(8)^2 + 10(8) - 16 = 3(64) + 80 - 16 = 192 + 80 - 16 = 256 > 0$$$ Base: $2(8) = 16 > 0$ and $16 \neq 1$, valid. - For $x=2$: $$$3(2)^2 + 10(2) - 16 = 3(4) + 20 - 16 = 12 + 20 - 16 = 16 > 0$$$ Base: $2(2) = 4 > 0$ and $4 \neq 1$, valid. 10. **Final answer:** $$$x = 2 \text{ or } x = 8$$$