What is the answer to 2^x+2^{4-x}=17 ?

The answer to 2^x+2^{4-x}=17 is x=4,x=0

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2^x+2^{4-x}=17

2^{x} + 2^{4 - x} = 17

x = 4, x = 0

Solution steps

2^{x} + 2^{4 - x} = 17

Apply exponent rules

2^{x} + (2^{x})^{- 1} \cdot 2^{4} = 17

Rewrite the equation with

2^{x} = u

u + (u)^{- 1} \cdot 2^{4} = 17

Solve

u + u^{- 1} \cdot 2^{4} = 17 : u = 16, u = 1

u = 16, u = 1

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 16 : x = 4

Solve

2^{x} = 1 : x = 0

x = 4, x = 0

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

x^{x} = (x)^{x}

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

5^{3 r - 5} = 5^{2}

0 = e^{x^{2}}