determine all the values of x, such that (x^2-15)^x^2-4 = 1.
We know (any base)^0 = 1
so x^2 - 4 = 0
x = ± 2
Also, 1 to the power of any finite number is also 1.
So can you find the remaining solutions?
If x = 4 or -4, x^2 -15 = 1, so no matter what x^2-4 is, the function
(x^2-15)^x^2-4 = 1 equals 1 when x = + or - 4.
Therefore -4 and 4 are also solutions.
Thanks to MathMate for pointing that out.
To determine all the values of x that satisfy the equation (x^2-15)^(x^2-4) = 1, we need to solve the equation.
Firstly, we know that any number raised to the power of 0 is equal to 1, so the exponent on the left side of the equation must be equal to 0. So, we have two cases to consider:
Case 1: (x^2-15) = 0
To solve this equation, we set x^2-15 = 0 and solve for x:
x^2 - 15 = 0
x^2 = 15
Taking the square root of both sides, we have
x = ±√(15)
Therefore, the solutions for Case 1 are x = √(15) and x = -√(15).
Case 2: (x^2-4) = 0
To solve this equation, we set x^2-4 = 0 and solve for x:
x^2 - 4 = 0
x^2 = 4
Taking the square root of both sides, we have
x = ±√(4)
Therefore, the solutions for Case 2 are x = 2 and x = -2.
Therefore, the values of x that satisfy the given equation are x = √(15), x = -√(15), x = 2, and x = -2.