What is the answer to 3^x+3^{1-x}-4=0 ?

The answer to 3^x+3^{1-x}-4=0 is x=1,x=0

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3^x+3^{1-x}-4=0

3^{x} + 3^{1 - x} - 4 = 0

x = 1, x = 0

Solution steps

3^{x} + 3^{1 - x} - 4 = 0

Apply exponent rules

3^{x} + (3^{x})^{- 1} \cdot 3^{1} - 4 = 0

Rewrite the equation with

3^{x} = u

u + (u)^{- 1} \cdot 3^{1} - 4 = 0

Solve

u + u^{- 1} \cdot 3^{1} - 4 = 0 : u = 3, u = 1

u = 3, u = 1

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 3 : x = 1

Solve

3^{x} = 1 : x = 0

x = 1, x = 0

2^{x - 1} - 2^{x - 3} = 3^{x - 2} - 3^{x - 3}

2^{x + 1} + 2^{3 - x} = 17

2^{5 x} = 94

\frac{1}{8} = 8^{x}

e^{x} - 5 e^{- x} = 4

What is the answer to 3^x+3^{1-x}-4=0 ?
The answer to 3^x+3^{1-x}-4=0 is x=1,x=0
Solve for x: 3^x+3^{1-x}-4=0
The solution is x=1,x=0