What is the answer to 2^{2x-1}-5*2^{x+1}+32=0 ?

The answer to 2^{2x-1}-5*2^{x+1}+32=0 is x=4,x=2

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2^{2x-1}-5*2^{x+1}+32=0

2^{2 x - 1} - 5 \cdot 2^{x + 1} + 32 = 0

x = 4, x = 2

Solution steps

2^{2 x - 1} - 5 \cdot 2^{x + 1} + 32 = 0

Apply exponent rules

(2^{x})^{2} \cdot 2^{- 1} - 5 \cdot 2^{x} \cdot 2^{1} + 32 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} \cdot 2^{- 1} - 5 u \cdot 2^{1} + 32 = 0

Solve

u^{2} \cdot 2^{- 1} - 5 u \cdot 2^{1} + 32 = 0 : u = 16, u = 4

u = 16, u = 4

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 16 : x = 4

Solve

2^{x} = 4 : x = 2

x = 4, x = 2

1^{x} = 0

1^{x} = 1

4^{z} = 8

2^{x} = 4096

3^{x + 2} + 3^{x} = 2430

What is the answer to 2^{2x-1}-5*2^{x+1}+32=0 ?
The answer to 2^{2x-1}-5*2^{x+1}+32=0 is x=4,x=2
Solve for x: 2^{2x-1}-5*2^{x+1}+32=0
The solution is x=4,x=2