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

The answer to 2^x-2^{-x}=5(1-2^{-x}) is x=2,x=0

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

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

x = 2, x = 0

Solution steps

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

Apply exponent rules

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

Rewrite the equation with

2^{x} = u

u - (u)^{- 1} = 5 (1 - (u)^{- 1})

Solve

u - u^{- 1} = 5 (1 - u^{- 1}) : u = 4, u = 1

u = 4, u = 1

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 4 : x = 2

Solve

2^{x} = 1 : x = 0

x = 2, x = 0

3^{x - 2} \cdot 2^{x - 1} = 1125

4^{x} = (2^{x})^{2}

2^{(x)} + 4 x + 12 = 0

e^{- x} + 3 = 0

3 6^{\frac{1}{2} - 4 x} = 6

What is the answer to 2^x-2^{-x}=5(1-2^{-x}) ?
The answer to 2^x-2^{-x}=5(1-2^{-x}) is x=2,x=0
Solve for x: 2^x-2^{-x}=5(1-2^{-x})
The solution is x=2,x=0