What is the answer to 2^{2x}-12*2^x+32=0 ?

The answer to 2^{2x}-12*2^x+32=0 is x=3,x=2

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2^{2x}-12*2^x+32=0

2^{2 x} - 12 \cdot 2^{x} + 32 = 0

x = 3, x = 2

Solution steps

2^{2 x} - 12 \cdot 2^{x} + 32 = 0

Apply exponent rules

(2^{x})^{2} - 12 \cdot 2^{x} + 32 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} - 12 u + 32 = 0

Solve

u^{2} - 12 u + 32 = 0 : u = 8, u = 4

u = 8, u = 4

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 8 : x = 3

Solve

2^{x} = 4 : x = 2

x = 3, x = 2

1 0^{5 + 8 x} = 79800

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

2^{2 x + 1} - 5 (2^{x}) - 12 = 0

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

2. 5^{- 4 r} = 6.4

What is the answer to 2^{2x}-12*2^x+32=0 ?
The answer to 2^{2x}-12*2^x+32=0 is x=3,x=2
Solve for x: 2^{2x}-12*2^x+32=0
The solution is x=3,x=2