What is the answer to ((2^{1-x}-2^x+1))/((2^x-1))=0 ?

The answer to ((2^{1-x}-2^x+1))/((2^x-1))=0 is x=1

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((2^{1-x}-2^x+1))/((2^x-1))=0

\frac{( 2 ^{1 - x} - 2 ^{x} + 1 )}{( 2 ^{x} - 1 )} = 0

x = 1

Solution steps

\frac{( 2 ^{1 - x} - 2 ^{x} + 1 )}{( 2 ^{x} - 1 )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2^{1 - x} - 2^{x} + 1 = 0

Apply exponent rules

(2^{x})^{- 1} \cdot 2^{1} - 2^{x} + 1 = 0

Rewrite the equation with

2^{x} = u

(u)^{- 1} \cdot 2^{1} - u + 1 = 0

Solve

u^{- 1} \cdot 2^{1} - u + 1 = 0 : u = - 1, u = 2

u = - 1, u = 2

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = - 1 :

No Solution for

x \in R

Solve

2^{x} = 2 : x = 1

x = 1

Verify Solutions:

x = 1

True

The solution is

x = 1

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

9^{n + 1} - 2 \cdot 9^{n} = 7

\frac{8}{3} = 2^{r}

\frac{4 ^{x + 2}}{3 ^{x + 1}} = 16

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

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