What is the answer to (2\times (-5))^{x^2-5x+4}=1 ?

The answer to (2\times (-5))^{x^2-5x+4}=1 is x=1,x=4

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(2\times (-5))^{x^2-5x+4}=1

(2 \cdot (- 5))^{x^{2} - 5 x + 4} = 1

x = 1, x = 4

Solution steps

(2 (- 5))^{x^{2} - 5 x + 4} = 1

(- a)^{n} = b,

assuming

n

is even, solve

a^{n} = b

instead

Solve

1 0^{x^{2} - 5 x + 4} = 1 : x = 1, x = 4

Verify Solutions:

x = 1

True

, x = 4

True

The solutions are

x = 1, x = 4

x^{x} = 6

3^{3 x + 5} = 3 \cdot 9^{x + 3}

9^{x} - 10 \times 3^{x} + 9 = 0

5^{a + 2} = \frac{1}{125}

x, y = 1 2^{1 - 0.1 x}

What is the answer to (2\times (-5))^{x^2-5x+4}=1 ?
The answer to (2\times (-5))^{x^2-5x+4}=1 is x=1,x=4
Solve for x: (2\times (-5))^{x^2-5x+4}=1
The solution is x=1,x=4