This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Step 1: Solve equation A. Equation A is 4^(2x^2+x-6) = 1. For an equation of the form a^b = 1, if a ≠ 1 and a ≠ -1, then the exponent b must be 0. Here, the base is 4, so we set the exponent to 0: 2x^2+x-6 = 0 We can solve this quadratic equation by factoring. We look for two numbers that multiply to 2 × (-6) = -12 and add to 1. These numbers are 4 and -3. 2x^2+4x-3x-6 = 0 2x(x+2)-3(x+2) = 0 (2x-3)(x+2) = 0 Setting each factor to zero: 2x-3 = 0 2x = 3 x = (3)/(2) x+2 = 0 x = -2 a) Solutions for equation A: x = (3)/(2), -2 Step 2: Solve equation B. Equation B is (x^2-3x+1)^6 = 1. For an equation of the form B^E = 1, where E is an even integer (here E=6), the base B can be 1 or -1. Case 1: The base is 1. x^2-3x+1 = 1 x^2-3x = 0 x(x-3) = 0 This gives x=0 or x=3. Case 2: The base is -1. x^2-3x+1 = -1 x^2-3x+2 = 0 Factoring the quadratic: (x-1)(x-2) = 0 This gives x=1 or x=2. b) Solutions for equation B: x = 0, 1, 2, 3 Step 3: Solve equation C. Equation C is (x^2-3x+1)^(2x^2+x-6) = 1. Let the base be B = x^2-3x+1 and the exponent be E = 2x^2+x-6. We need to solve B^E = 1. There are three cases: Case 1: The exponent is 0, provided the base is not 0. E = 2x^2+x-6 = 0 From Step 1, the solutions are x = (3)/(2) and x = -2. We check the base B for these values: For x = (3)/(2): B = ((3)/(2))^2 - 3((3)/(2)) + 1 = (9)/(4) - (9)/(2) + 1 = (9-18+4)/(4) = -(5)/(4). Since B ≠ 0, x = (3)/(2) is a solution. For x = -2: B = (-2)^2 - 3(-2) + 1 = 4 + 6 + 1 = 11. Since B ≠ 0, x = -2 is a solution. Case 2: The base is 1. B = x^2-3x+1 = 1 From Step 2, the solutions are x = 0 and x = 3. For these values, 1^E = 1 holds for any real exponent E. So, x=0 and x=3 are solutions. Case 3: The base is -1, and the exponent is an even integer. B = x^2-3x+1 = -1 From Step 2, the solutions are x = 1 and x = 2. We check if the exponent E is even for these values: For x = 1: E = 2(1)^2 + (1) - 6 = 2 + 1 - 6 = -3. Since -3 is an odd integer, x=1 is not a solution. For x = 2: E = 2(2)^2 + (2) - 6 = 8 + 2 - 6 = 4. Since 4 is an even integer, x=2 is a solution. Combining all valid solutions from the three cases: x = -2, 0, (3)/(2), 2, 3 c) Solutions for equation C: x = -2, 0, (3)/(2), 2, 3 Step 4: State how many values of x satisfy each equation. a) Equation A: From Step 1, there are 2 solutions: x = (3)/(2), -2. Number of values: 2 b) Equation B: From Step 2, there are 4 solutions: x = 0, 1, 2, 3. Number of values: 4 c) Equation C: From Step 3, there are 5 solutions: x = -2, 0, (3)/(2), 2, 3. Number of values: 5