1. Problem 5: Solve the equation $$2^{x-1} + 2^x + 2^{x+1} = 7$$ 2. Use the property of exponents: $$2^{x-1} = \frac{2^x}{2}$$ and $$2^{x+1} = 2 \cdot 2^x$$. 3. Rewrite the equation as $$\frac{2^x}{2} + 2^x + 2 \cdot 2^x = 7$$. 4. Let $$y = 2^x$$, then the equation becomes $$\frac{y}{2} + y + 2y = 7$$. 5. Combine like terms: $$\frac{y}{2} + y + 2y = \frac{y}{2} + \frac{2y}{2} + \frac{4y}{2} = \frac{7y}{2}$$. 6. So, $$\frac{7y}{2} = 7$$ which implies $$y = 2$$. 7. Recall $$y = 2^x$$, so $$2^x = 2$$, giving $$x = 1$$. 1. Problem 6: Solve $$5^{x+2} + 5^{x+1} + 5^x = 31$$. 2. Use exponent properties: $$5^{x+2} = 5^2 \cdot 5^x = 25 \cdot 5^x$$ and $$5^{x+1} = 5 \cdot 5^x$$. 3. Let $$y = 5^x$$, then equation becomes $$25y + 5y + y = 31$$. 4. Combine terms: $$31y = 31$$, so $$y = 1$$. 5. Since $$y = 5^x$$, $$5^x = 1$$ implies $$x = 0$$. 1. Problem 7: Solve $$x - \sqrt{2x - 4} = 6$$. 2. Isolate the square root: $$\sqrt{2x - 4} = x - 6$$. 3. Square both sides: $$2x - 4 = (x - 6)^2 = x^2 - 12x + 36$$. 4. Rearrange: $$0 = x^2 - 12x + 36 - 2x + 4 = x^2 - 14x + 40$$. 5. Solve quadratic: $$x^2 - 14x + 40 = 0$$. 6. Use quadratic formula: $$x = \frac{14 \pm \sqrt{14^2 - 4 \cdot 40}}{2} = \frac{14 \pm \sqrt{196 - 160}}{2} = \frac{14 \pm \sqrt{36}}{2}$$. 7. So, $$x = \frac{14 \pm 6}{2}$$, giving $$x = 10$$ or $$x = 4$$. 8. Check solutions in original equation: - For $$x=10$$: $$10 - \sqrt{20 - 4} = 10 - \sqrt{16} = 10 - 4 = 6$$ (valid). - For $$x=4$$: $$4 - \sqrt{8 - 4} = 4 - \sqrt{4} = 4 - 2 = 2 \neq 6$$ (invalid). 9. Final solution: $$x = 10$$. 1. Problem 8: Solve $$\frac{2}{3} + \frac{x(2x + 3)}{4} - \frac{x(x - 3)}{6} = 1$$. 2. Find common denominator 12 and rewrite: $$\frac{8}{12} + \frac{3x(2x + 3)}{12} - \frac{2x(x - 3)}{12} = 1$$. 3. Combine terms: $$\frac{8 + 3x(2x + 3) - 2x(x - 3)}{12} = 1$$. 4. Multiply both sides by 12: $$8 + 3x(2x + 3) - 2x(x - 3) = 12$$. 5. Expand: $$8 + 6x^2 + 9x - 2x^2 + 6x = 12$$. 6. Simplify: $$8 + 4x^2 + 15x = 12$$. 7. Rearrange: $$4x^2 + 15x + 8 - 12 = 0 \Rightarrow 4x^2 + 15x - 4 = 0$$. 8. Use quadratic formula: $$x = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 4 \cdot (-4)}}{2 \cdot 4} = \frac{-15 \pm \sqrt{225 + 64}}{8} = \frac{-15 \pm \sqrt{289}}{8} = \frac{-15 \pm 17}{8}$$. 9. Solutions: $$x = \frac{2}{8} = \frac{1}{4}$$ or $$x = \frac{-32}{8} = -4$$. 1. Problem 9: Solve $$\frac{x(3x + 2)}{3} - \frac{7x^2}{6} = x - \frac{3x + 1}{12}$$. 2. Multiply both sides by 12 to clear denominators: $$4x(3x + 2) - 14x^2 = 12x - (3x + 1)$$. 3. Expand: $$12x^2 + 8x - 14x^2 = 12x - 3x - 1$$. 4. Simplify: $$-2x^2 + 8x = 9x - 1$$. 5. Rearrange: $$-2x^2 + 8x - 9x + 1 = 0 \Rightarrow -2x^2 - x + 1 = 0$$. 6. Multiply by -1: $$2x^2 + x - 1 = 0$$. 7. Use quadratic formula: $$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$. 8. Solutions: $$x = \frac{2}{4} = \frac{1}{2}$$ or $$x = \frac{-4}{4} = -1$$. Final answers: - Problem 5: $$x=1$$ - Problem 6: $$x=0$$ - Problem 7: $$x=10$$ - Problem 8: $$x=\frac{1}{4}, -4$$ - Problem 9: $$x=\frac{1}{2}, -1$$