4 Find the real solutions of the following equations. C 2^(x^(2-4x+6))=8 a 8^(x^(2+2x-15))=1 4^(2x^(2-11x+15))=1 d 3^(2x^(2+9x+2))=(1)/(9) e (x^2+2x-14)^5=1 f (x^2-7x+11)^8=1

Let's solve each equation step-by-step:**c) 2(x² - 4x + 6) = 8**Since 8 = 2³, we can rewrite the equation as:2(x² - 4x + 6) = 2³The bases are the same, so we can equate the exponents:x² - 4x + 6 = 3x² - 4x + 3 = 0This is a quadratic equation. We can factor it:(x - 1)(x - 3) = 0Therefore, the real solutions are x = 1 and x = 3.**a) 8(x² + 2x - 15) = 1**Since any number raised to the power of 0 equals 1, we have:x² + 2x - 15 = 0This is a quadratic equation. We can factor it:(x + 5)(x - 3) = 0Therefore, the real solutions are x = -5 and x = 3.**d) 4(2x² - 11x + 15) = 1**Similar to part (a), we have:2x² - 11x + 15 = 0This quadratic equation can be factored:(2x - 5)(x - 3) = 0Therefore, the real solutions are x = 5/2 and x = 3.**d) 3(2x² + 9x + 2) = 1/9**Since 1/9 = 3⁻², we can rewrite the equation as:3(2x² + 9x + 2) = 3⁻²Equating the exponents:2x² + 9x + 2 = -22x² + 9x + 4 = 0This quadratic equation can be factored:(2x + 1)(x + 4) = 0Therefore, the real solutions are x = -1/2 and x = -4.**e) (x² + 2x - 14)⁵ = 1**Taking the fifth root of both sides:x² + 2x - 14 = 1x² + 2x - 15 = 0(x + 5)(x - 3) = 0Therefore, the real solutions are x = -5 and x = 3.**f) (x² - 7x + 11)⁸ = 1**Taking the eighth root of both sides:x² - 7x + 11 = ±1This gives us two equations:1) x² - 7x + 11 = 1 => x² - 7x + 10 = 0 => (x - 2)(x - 5) = 0 => x = 2, x = 52) x² - 7x + 11 = -1 => x² - 7x + 12 = 0 => (x - 3)(x - 4) = 0 => x = 3, x = 4Therefore, the real solutions are x = 2, x = 3, x = 4, and x = 5.