What is the answer to 2^0+2^{1-2x}=9*2^{-1-x} ?

The answer to 2^0+2^{1-2x}=9*2^{-1-x} is x=2,x=-1

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2^0+2^{1-2x}=9*2^{-1-x}

2^{0} + 2^{1 - 2 x} = 9 \cdot 2^{- 1 - x}

x = 2, x = - 1

Solution steps

2^{0} + 2^{1 - 2 x} = 9 \cdot 2^{- 1 - x}

Apply exponent rules

2^{0} + (2^{x})^{- 2} \cdot 2^{1} = 9 (2^{x})^{- 1} \cdot 2^{- 1}

Rewrite the equation with

2^{x} = u

2^{0} + (u)^{- 2} \cdot 2^{1} = 9 (u)^{- 1} \cdot 2^{- 1}

Solve

2^{0} + u^{- 2} \cdot 2^{1} = 9 u^{- 1} \cdot 2^{- 1} : u = 4, u = \frac{1}{2}

u = 4, u = \frac{1}{2}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 4 : x = 2

Solve

2^{x} = \frac{1}{2} : x = - 1

x = 2, x = - 1

x, U^{y} Ux = e^{x} U y

1^{x} = 4

1^{x} = 3

x, f = (\frac{1}{3})^{x}

x^{2 l o g_{3} (x)} = 9

What is the answer to 2^0+2^{1-2x}=9*2^{-1-x} ?
The answer to 2^0+2^{1-2x}=9*2^{-1-x} is x=2,x=-1
Solve for x: 2^0+2^{1-2x}=9*2^{-1-x}
The solution is x=2,x=-1