What is the answer to 2^{2x+1}-33(2^{x-1})+4=0 ?

The answer to 2^{2x+1}-33(2^{x-1})+4=0 is x=3,x=-2

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2^{2x+1}-33(2^{x-1})+4=0

2^{2 x + 1} - 33 (2^{x - 1}) + 4 = 0

x = 3, x = - 2

Solution steps

2^{2 x + 1} - 33 (2^{x - 1}) + 4 = 0

Apply exponent rules

(2^{x})^{2} \cdot 2^{1} - 33 \cdot 2^{x} \cdot 2^{- 1} + 4 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} \cdot 2^{1} - 33 u \cdot 2^{- 1} + 4 = 0

Solve

u^{2} \cdot 2^{1} - 33 u \cdot 2^{- 1} + 4 = 0 : u = 8, u = \frac{1}{4}

u = 8, u = \frac{1}{4}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 8 : x = 3

Solve

2^{x} = \frac{1}{4} : x = - 2

x = 3, x = - 2

5^{x - 3} + 12 5^{x + 3} = 78125

2^{3000} = 1 0^{x}

(\frac{1}{2})^{n} + 1 = 2

t, 4 \frac{( x ^{2} ) ^{- t}}{5} = 288

2^{x} + 3 \cdot 2^{x} = 16

What is the answer to 2^{2x+1}-33(2^{x-1})+4=0 ?
The answer to 2^{2x+1}-33(2^{x-1})+4=0 is x=3,x=-2
Solve for x: 2^{2x+1}-33(2^{x-1})+4=0
The solution is x=3,x=-2