What is the answer to 2^{2x}-5(2^x)+4=0 ?

The answer to 2^{2x}-5(2^x)+4=0 is x=2,x=0

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2^{2x}-5(2^x)+4=0

2^{2 x} - 5 (2^{x}) + 4 = 0

x = 2, x = 0

Solution steps

2^{2 x} - 5 (2^{x}) + 4 = 0

Apply exponent rules

(2^{x})^{2} - 5 \cdot 2^{x} + 4 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} - 5 u + 4 = 0

Solve

u^{2} - 5 u + 4 = 0 : u = 4, u = 1

u = 4, u = 1

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 4 : x = 2

Solve

2^{x} = 1 : x = 0

x = 2, x = 0

e^{- 3} e^{2} ln (e^{e}) = e^{x}

5^{x^{2}} = 17

x, 0 = e^{x}

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

What is the answer to 2^{2x}-5(2^x)+4=0 ?
The answer to 2^{2x}-5(2^x)+4=0 is x=2,x=0
Solve for x: 2^{2x}-5(2^x)+4=0
The solution is x=2,x=0