(x^2-5x+5)^(x^2+4x-60)=1

3 min read Jun 17, 2024

Solving the Equation (x^2-5x+5)^(x^2+4x-60) = 1

This equation presents an interesting challenge because it involves an exponential expression with a variable in both the base and the exponent. To solve it, we'll use the following key concepts:

Understanding the Properties of Exponents

Anything raised to the power of 0 equals 1: If the exponent (x^2+4x-60) equals 0, the entire expression becomes 1.
A number raised to the power of 1 equals itself: If the base (x^2-5x+5) equals 1, the entire expression becomes 1.

Solving for the Exponent

1. Solve the equation x^2+4x-60 = 0:

This is a quadratic equation, which can be factored: (x+10)(x-6) = 0 Therefore, x = -10 or x = 6.

Solving for the Base

2. Solve the equation x^2-5x+5 = 1:

Subtract 1 from both sides to get: x^2 - 5x + 4 = 0 Factor the equation: (x-4)(x-1) = 0 Therefore, x = 4 or x = 1.

Combining the Solutions

We've found four possible values of x that could satisfy the original equation:

x = -10
x = 6
x = 4
x = 1

Important Note: It's crucial to check each solution by plugging it back into the original equation to ensure it's valid. There may be extraneous solutions that arise from our algebraic manipulations.

Checking the solutions:

x = -10: ((-10)^2 - 5(-10) + 5)^((-10)^2 + 4(-10) - 60) = (155)^0 = 1 (Valid solution)
x = 6: (6^2 - 5(6) + 5)^(6^2 + 4(6) - 60) = 1^0 = 1 (Valid solution)
x = 4: (4^2 - 5(4) + 5)^(4^2 + 4(4) - 60) = 1^0 = 1 (Valid solution)
x = 1: (1^2 - 5(1) + 5)^(1^2 + 4(1) - 60) = 1^(-55) = 1 (Valid solution)

Therefore, all four solutions are valid: x = -10, x = 6, x = 4, and x = 1.