What is the answer to (4-x)^{x^3-4x}=(4-x)^{(x^2-4)(x-1)} ?

The answer to (4-x)^{x^3-4x}=(4-x)^{(x^2-4)(x-1)} is x=3,x=2,x=-2

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(4-x)^{x^3-4x}=(4-x)^{(x^2-4)(x-1)}

(4 - x)^{x^{3} - 4 x} = (4 - x)^{(x^{2} - 4) (x - 1)}

x = 3, x = 2, x = - 2

Solution steps

(4 - x)^{x^{3} - 4 x} = (4 - x)^{(x^{2} - 4) (x - 1)}

Apply exponent rules

(x^{3} - 4 x) ln (4 - x) = (x^{2} - 4) (x - 1) ln (4 - x)

Solve

(x^{3} - 4 x) ln (4 - x) = (x^{2} - 4) (x - 1) ln (4 - x) : x = 3, x = 2, x = - 2

x = 3, x = 2, x = - 2

6^{x} = 9^{x}

2 7^{x - 1} = (\frac{1}{9})^{x}

x, 2 u + 3 v = e^{x} (cos (y) - sin (y))

e^{y} = 0

2^{x + 3} + 2^{x} = 72

What is the answer to (4-x)^{x^3-4x}=(4-x)^{(x^2-4)(x-1)} ?
The answer to (4-x)^{x^3-4x}=(4-x)^{(x^2-4)(x-1)} is x=3,x=2,x=-2
Solve for x: (4-x)^{x^3-4x}=(4-x)^{(x^2-4)(x-1)}
The solution is x=3,x=2,x=-2