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

The answer to (2^{x^2-2x})^{4-x}=1 is x=4,x=0,x=2

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

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

x = 4, x = 0, x = 2

Solution steps

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

Apply exponent rules

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

4 - x = 0 or ln (2^{x^{2} - 2 x}) = 0

Solve

4 - x = 0 : x = 4

Solve

ln (2^{x^{2} - 2 x}) = 0 : x = 0, x = 2

x = 4, x = 0, x = 2

100 0^{x} = 10

2 \times 5^{x} = 250

4 9^{p + 1} = 34 3^{2 p}

0.9 8^{x} = 1.6

What is the answer to (2^{x^2-2x})^{4-x}=1 ?
The answer to (2^{x^2-2x})^{4-x}=1 is x=4,x=0,x=2
Solve for x: (2^{x^2-2x})^{4-x}=1
The solution is x=4,x=0,x=2