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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Welcome back MEDIC — missed you this week. Here are the solutions for each equation: a) Solve 2^2x+3 - 9 × 2^x+1 + 1 = 0 Step 1: Rewrite the equation using exponent rules a^m+n = a^m · a^n and a^mn = (a^m)^n. 2^2x · 2^3 - 9 · 2^x · 2^1 + 1 = 0 8 · (2^x)^2 - 18 · 2^x + 1 = 0 Step 2: Let y = 2^x. Substitute y into the equation to form a quadratic equation. 8y^2 - 18y + 1 = 0 Step 3: Solve the quadratic equation for y using the quadratic formula y = -b ± sqrt(b^2 - 4ac)2a. Here, a=8, b=-18, c=1. y = -(-18) ± sqrt((-18)^2 - 4(8)(1))2(8) y = 18 ± sqrt(324 - 32)16 y = 18 ± sqrt(292)16 y = 18 ± 2sqrt(73)16 y = 9 ± sqrt(73)8 Step 4: Substitute back y = 2^x and solve for x. 2^x = 9 + sqrt(73)8 or 2^x = 9 - sqrt(73)8 Take the logarithm base 2 on both sides. x = _2(9 + sqrt(73)8) or x = _2(9 - sqrt(73)8) Since sqrt(73) ≈ 8.54, 9 - sqrt(73)8 ≈ (9 - 8.54)/(8) = (0.46)/(8) > 0. Both solutions are valid. The solutions are x = _2(9 + sqrt(73)8), _2(9 - sqrt(73)8). b) Solve 3^x × 3^x - 28 × 3^x-1 + 3 = 0 Step 1: Rewrite the equation using exponent rules (a^m)^n = a^mn and a^m-n = (a^m)/(a^n). (3^x)^2 - 28 × (3^x)/(3^1) + 3 = 0 (3^x)^2 - (28)/(3) · 3^x + 3 = 0 Step 2: Let y = 3^x. Substitute y into the equation. y^2 - (28)/(3)y + 3 = 0 Multiply by 3 to clear the fraction: 3y^2 - 28y + 9 = 0 Step 3: Solve the quadratic equation for y by factoring. We look for two numbers that multiply to 3 × 9 = 27 and add to -28. These numbers are -27 and -1. 3y^2 - 27y - y + 9 = 0 3y(y - 9) - 1(y - 9) = 0 (3y - 1)(y - 9) = 0 So, 3y - 1 = 0 y = (1)/(3) or y - 9 = 0 y = 9. Step 4: Substitute back y = 3^x and solve for x. For y = (1)/(3): 3^x = (1)/(3) 3^x = 3^-1 x = -1 For y = 9: 3^x = 9 3^x = 3^2 x = 2 The solutions are x = -1, 2. c) Solve 2^2x-1 - 5 × 2^x-1 = 12 Step 1: Rewrite the equation using exponent rules a^m-n = (a^m)/(a^n) and a^mn = (a^m)^n. 2^2x2^1 - 5 × (2^x)/(2^1) = 12 ((2^x)^2)/(2) - (5 · 2^x)/(2) = 12 Step 2: Let y = 2^x. Substitute y into the equation. (y^2)/(2) - (5y)/(2) = 12 Multiply by 2 to clear the fractions: y^2 - 5y = 24 Step 3: Rearrange the equation into standard quadratic form ay^2 + by + c = 0. y^2 - 5y - 24 = 0 Step 4: Solve the quadratic equation for y by factoring. We look for two numbers that multiply to -24 and add to -5. These numbers are -8 and 3. (y - 8)(y + 3) = 0 So, y - 8 = 0 y = 8 or y + 3 = 0 y = -3. Step 5: Substitute back y = 2^x and solve for x. For y = 8: 2^x = 8 2^x = 2^3 x = 3 For y = -3: 2^x = -3 This equation has no real solution, as 2^x is always positive. The solution is x = 3. d) Solve 5^x+3 - 5 ÷ 5^x + 120 = 0 Step 1: Rewrite the equation using exponent rules a^m+n = a^m · a^n and a^m ÷ a^n = a^m-n or (a^m)/(a^n). 5^x · 5^3 - (5)/(5^x) + 120 = 0 125 · 5^x - (5)/(5^x) + 120 = 0 Step 2: Let y = 5^x. Substitute y into the equation. 125y - (5)/(y) + 120 = 0 Step 3: Multiply the entire equation by y to eliminate the fraction. Note that y = 5^x must be positive, so y ≠ 0. 125y^2 - 5 + 120y = 0 Step 4: Rearrange the equation into standard quadratic form ay^2 + by + c = 0. 125y^2 + 120y - 5 = 0 Divide by 5 to simplify: 25y^2 + 24y - 1 = 0 Step 5: Solve the quadratic equation for y by factoring. We look for two numbers that multiply to 25 × (-1) = -25 and add to 24. These numbers are 25 and -1. 25y^2 + 25y - y - 1 = 0 25y(y + 1) - 1(y + 1) = 0 (25y - 1)(y + 1) = 0 So, 25y - 1 = 0 y = (1)/(25) or y + 1 = 0 y = -1. Step 6: Substitute back y = 5^x and solve for x. For y = (1)/(25): 5^x = (1)/(25) 5^x = 5^-2 x = -2 For y = -1: 5^x = -1 This equation has no real solution, as 5^x is always positive. The solution is x = -2. e) Solve 3^x × 3^x - 10 × 3^x-1 + 1 = 0 Step 1: Rewrite the equation using exponent rules (a^m)^n = a^mn and a^m-n = (a^m)/(a^n). (3^x)^2 - 10 × (3^x)/(3^1) + 1 = 0 (3^x)^2 - (10)/(3) · 3^x + 1 = 0 Step 2: Let y = 3^x. Substitute y into the equation. y^2 - (10)/(3)y + 1 = 0 Multiply by 3 to clear the fraction: 3y^2 - 10y + 3 = 0 Step 3: Solve the quadratic equation for y by factoring. We look for two numbers that multiply to 3 × 3 = 9 and add to -10. These numbers are -9 and -1. 3y^2 - 9y - y + 3 = 0 3y(y - 3) - 1(y - 3) = 0 (3y - 1)(y - 3) = 0 So, 3y - 1 = 0 y = (1)/(3) or y - 3 = 0 y = 3. Step 4: Substitute back y = 3^x and solve for x. For y = (1)/(3): 3^x = (1)/(3) 3^x = 3^-1 x = -1 For y = 3: 3^x = 3 3^x = 3^1 x = 1 The solutions are x = -1, 1. Drop the next question!