What is the answer to 2(2^{2x})-5(2^x)+2=0 ?

The answer to 2(2^{2x})-5(2^x)+2=0 is x=1,x=-1

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2(2^{2x})-5(2^x)+2=0

2 (2^{2 x}) - 5 (2^{x}) + 2 = 0

x = 1, x = - 1

Solution steps

2 (2^{2 x}) - 5 (2^{x}) + 2 = 0

Apply exponent rules

(2^{x})^{2} \cdot 2^{1} - 5 \cdot 2^{x} + 2 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} \cdot 2^{1} - 5 u + 2 = 0

Solve

u^{2} \cdot 2^{1} - 5 u + 2 = 0 : u = 2, u = \frac{1}{2}

u = 2, u = \frac{1}{2}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 2 : x = 1

Solve

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

x = 1, x = - 1

2^{x + 1} + 2^{x} = 48

4 e^{3 x} = 9

x, 2^{x} + 8^{y} = 16

4^{x} = 100

2 7^{9 x + 4} \cdot 3^{x - 8} = 3^{x + 9}

What is the answer to 2(2^{2x})-5(2^x)+2=0 ?
The answer to 2(2^{2x})-5(2^x)+2=0 is x=1,x=-1
Solve for x: 2(2^{2x})-5(2^x)+2=0
The solution is x=1,x=-1