Solve and explain this exponential equation, please?? 2^((x+1)/x) * (1/2)^x+1= 1 The results are -1 and 1.?

2^[(x + 1)/x] * (1/2)^(x + 1) = 1

2^[(x + 1)/x] * [2^(-1)]^(x + 1) = 2^0

2^[(x + 1)/x] * 2^[-(x + 1)] = 2^0

2^[(x + 1)/x] * 2^[-(x² + x)/x] = 2^0

2^{[(x + 1)/x] + [-(x² + x)/x]} = 2^0

[(x + 1)/x] - [(x² + x)/x] = 0

(x + 1)/x = (x² + x)/x

For x ≠ 0, x + 1 = x² + x

x² - 1 = 0

(x + 1)(x - 1) = 0

x = -1 or x = 1

I think you forgot parentheses around that last exponent.

2^((x+1)/x)·(½)ˣ⁺¹ = 1
2^((x+1)/x) = (½)⁻⁽ˣ⁺¹⁾
2^((x+1)/x) = 2⁽ˣ⁺¹⁾
(x+1)/x = x+1
(x+1) = x(x+1)
x = 1 or x+1 = 0
x = 1 or -1

2^((x + 1)/x) * (2^-1)^(x + 1) = 1
2^((x + 1)/x) * 2^(-1(x + 1)) = 1
2^((x + 1)/x) * 2^(-x - 1) = 1
2^((x + 1)/x + -x - 1) = 1
(x + 1)/x - x - 1 = 0
(x + 1)/x = x + 1
x + 1 = x^2 + x
0 = x^2 - 1
x = 1 or x = -1

2^((x+1)/x) * (1/2)^x+1= 1

did you mean

2^((x+1)/x) * (1/2)^(x+1) = 1 ???

2^((x+1)/x) * (2^(-1))^(x+1) = 1

2^((x+1)/x) * 2^(-x-1) = 1 <<< multiplying == add exponents

2^[((x+1)/x)+(-x-1)] = 1 <<<< remember that ANYTHING raised to the 0 power = 1 .... 5^0 = 1 ... x^0 = 1 ... but use 2^0 to sub for 1

2^[((x+1)/x)+(-x-1)] = 2^0 <<<< now you can use logs or an algebra rule ... if the bases are the same then exponent = exponent ... cancelling the bases ex: 3^a = 3^b then a = b

[((x+1)/x) - (x+1)] = 0 <<< base 2 cancels
[(x + 1) - x( x + 1)]/x = 0 ... multiply by x
x + 1 - x^2 - x = 0
x^2 - 1 = 0
(x + 1)(x - 1) = 0
x = 1 and x = -1

2^((x+1)/x -x-1) = 1
2^ (1+1/x-x-1) = 1
2^(1/x -x) = 1
2^ ((1-x^2) /x ) = 1
1-x^2 / x = 0
1-x^2 = 0
x^2 = 1
x = ±1