Solve the Exponential Equation | (6 - x)^(x^2 + 7x + 12) = 1 | Quadratic Equation

Solve the Exponential Equation (6 - x)^(x^2+7x+12)=1

Solution of this equation involve three different cases
case 1: 6 - x = 1
case 2: 6 - x = -1 when x^2+7x+12 = even number
case 3: x^2+7x+12 = 0 when 6 - x ≠ 0

[justabunga]

Here is a shortcut to solve this equation instead of separating the cases. We can change (6-x)^(x^2+7x+12)=1 as e^((x+3)(x+4)ln(|6-x|)=e^0. Usually, the question is supposed to change it as e^((x+3)(x+4)ln(6-x)) without the absolute values due to the domain in the base since it has to be greater than 0 but not equal to 1. The reason why there is an absolute value is because we want to extend all the possible values of x since the domain isn't mentioned in the question. Otherwise, we only can consider positive values in the base. Here, since the base are the same, we have set the exponents equal to each other. It's solvable that (x+3)(x+4)ln(|6-x|)=0, so x=-4, -3, 5, and 7. We can confirm that all the values all correct.