What is the answer to 2(2^{2x})-8(2^x)+8=0 ?

The answer to 2(2^{2x})-8(2^x)+8=0 is x=1

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2(2^{2x})-8(2^x)+8=0

2 (2^{2 x}) - 8 (2^{x}) + 8 = 0

x = 1

Solution steps

2 (2^{2 x}) - 8 (2^{x}) + 8 = 0

Apply exponent rules

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

Rewrite the equation with

2^{x} = u

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

Solve

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

u = 2

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 2 : x = 1

x = 1

e^{6 k} = 11

61 = 3 3^{x}

3^{2 x^{2}} = 3^{9 x + 5}

x^{l o g_{5} (x)} = 400

4300 = 7900 (1 - 0.04)^{x}

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