1. **State the problem:** Solve the equation $$\log_2(6x + 1) + \log_2(4x - 2) = 7$$ for $x$. 2. **Recall the logarithm property:** The sum of logarithms with the same base can be combined as the logarithm of the product: $$\log_b A + \log_b B = \log_b (A \times B)$$ 3. **Apply the property:** $$\log_2(6x + 1) + \log_2(4x - 2) = \log_2\big((6x + 1)(4x - 2)\big)$$ So the equation becomes: $$\log_2\big((6x + 1)(4x - 2)\big) = 7$$ 4. **Rewrite the logarithmic equation in exponential form:** $$\log_2(y) = 7 \implies y = 2^7 = 128$$ Therefore: $$(6x + 1)(4x - 2) = 128$$ 5. **Expand the left side:** $$6x \times 4x = 24x^2$$ $$6x \times (-2) = -12x$$ $$1 \times 4x = 4x$$ $$1 \times (-2) = -2$$ So: $$24x^2 - 12x + 4x - 2 = 128$$ Simplify the middle terms: $$24x^2 - 8x - 2 = 128$$ 6. **Bring all terms to one side:** $$24x^2 - 8x - 2 - 128 = 0$$ $$24x^2 - 8x - 130 = 0$$ 7. **Simplify the quadratic equation by dividing all terms by 2:** $$\cancel{2} \times 12x^2 - \cancel{2} \times 4x - \cancel{2} \times 65 = 0$$ $$12x^2 - 4x - 65 = 0$$ 8. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=12$, $b=-4$, $c=-65$. Calculate the discriminant: $$\Delta = (-4)^2 - 4 \times 12 \times (-65) = 16 + 3120 = 3136$$ Calculate the square root: $$\sqrt{3136} = 56$$ 9. **Find the roots:** $$x = \frac{-(-4) \pm 56}{2 \times 12} = \frac{4 \pm 56}{24}$$ Two solutions: $$x_1 = \frac{4 + 56}{24} = \frac{60}{24} = \frac{5}{2} = 2.5$$ $$x_2 = \frac{4 - 56}{24} = \frac{-52}{24} = -\frac{13}{6} \approx -2.1667$$ 10. **Check domain restrictions:** - Inside the logarithms, arguments must be positive: $$6x + 1 > 0 \implies x > -\frac{1}{6}$$ $$4x - 2 > 0 \implies x > \frac{1}{2}$$ The stricter condition is $x > \frac{1}{2}$. 11. **Check solutions against domain:** - $x_1 = 2.5$ satisfies $x > 0.5$. - $x_2 \approx -2.1667$ does not satisfy $x > 0.5$. Therefore, the only valid solution is: $$\boxed{2.5}$$