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

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

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

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

x = 2, x = 3

Solution steps

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

Apply exponent rules

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

Solve

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

x = 2, x = 3

6 4^{a} = 8^{a + 2}

2^{2 x} = 4^{x + 1}

3^{x + 1} = 5^{2 x - 1}

2 e^{x + 1} = 18

x \cdot 2^{x} = 1

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