What is the answer to 2^{2x+3}-8*2^{x+1}+8=0 ?

The answer to 2^{2x+3}-8*2^{x+1}+8=0 is x=0

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2^{2x+3}-8*2^{x+1}+8=0

2^{2 x + 3} - 8 \cdot 2^{x + 1} + 8 = 0

x = 0

Solution steps

2^{2 x + 3} - 8 \cdot 2^{x + 1} + 8 = 0

Apply exponent rules

(2^{x})^{2} \cdot 2^{3} - 8 \cdot 2^{x} \cdot 2^{1} + 8 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} \cdot 2^{3} - 8 u \cdot 2^{1} + 8 = 0

Solve

u^{2} \cdot 2^{3} - 8 u \cdot 2^{1} + 8 = 0 : u = 1

u = 1

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 1 : x = 0

x = 0

2^{2 x - 5} = 8

7^{10} = 7^{5} \cdot 7^{f}

2^{x} - 5 = 59

4 e^{2 x} + e^{- x} = 0

y, x = 4^{y}

What is the answer to 2^{2x+3}-8*2^{x+1}+8=0 ?
The answer to 2^{2x+3}-8*2^{x+1}+8=0 is x=0
Solve for x: 2^{2x+3}-8*2^{x+1}+8=0
The solution is x=0