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

The answer to 4^x+4^{1-x}=4 is x= 1/2

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

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

x = \frac{1}{2}

Decimal Notation

x = 0.5

Solution steps

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

Apply exponent rules

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

Rewrite the equation with

4^{x} = u

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

Solve

u + u^{- 1} \cdot 4^{1} = 4 : u = 2

u = 2

Substitute back

u = 4^{x},

solve for

x

Solve

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

x = \frac{1}{2}

8^{x} \cdot 4^{x^{2} - 6} = 4

\frac{50}{4 + e ^{2 x}} = 11

5^{x^{2} - 2 x - 8} = 1

2^{x} = 2

4 (1 + 1 0^{5 x}) = 9