What is the answer to 3^x+3^{-x}=2 ?

The answer to 3^x+3^{-x}=2 is x=0

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

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

x = 0

Solution steps

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

Apply exponent rules

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

Rewrite the equation with

3^{x} = u

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

Solve

u + u^{- 1} = 2 : u = 1

u = 1

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 1 : x = 0

x = 0

x^{2} = 2^{x}

2^{x} = 5^{x + 2}

x^{x} = 2

3^{x} - 54 x + 135 = 0

2^{x^{2} - 4} = 3^{x^{2} - 4}