What is the answer to 8(2^{2x})-129(2^x)+16=0 ?

The answer to 8(2^{2x})-129(2^x)+16=0 is x=4,x=-3

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8(2^{2x})-129(2^x)+16=0

8 (2^{2 x}) - 129 (2^{x}) + 16 = 0

x = 4, x = - 3

Solution steps

8 (2^{2 x}) - 129 (2^{x}) + 16 = 0

Apply exponent rules

8 (2^{x})^{2} - 129 \cdot 2^{x} + 16 = 0

Rewrite the equation with

2^{x} = u

8 (u)^{2} - 129 u + 16 = 0

Solve

8 u^{2} - 129 u + 16 = 0 : u = 16, u = \frac{1}{8}

u = 16, u = \frac{1}{8}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 16 : x = 4

Solve

2^{x} = \frac{1}{8} : x = - 3

x = 4, x = - 3

2^{2 b - 3} = 32

2^{x - 4} = 7^{x}

3^{x + 2} = 5^{2 x + 7}

3^{2 x + 1} = 3^{π x + 4}

e^{x} + x - 1 = 0

What is the answer to 8(2^{2x})-129(2^x)+16=0 ?
The answer to 8(2^{2x})-129(2^x)+16=0 is x=4,x=-3
Solve for x: 8(2^{2x})-129(2^x)+16=0
The solution is x=4,x=-3