What is the answer to 64(4^{2x})-20(4^x)+1=0 ?

The answer to 64(4^{2x})-20(4^x)+1=0 is x=-1,x=-2

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64(4^{2x})-20(4^x)+1=0

64 (4^{2 x}) - 20 (4^{x}) + 1 = 0

x = - 1, x = - 2

Solution steps

64 (4^{2 x}) - 20 (4^{x}) + 1 = 0

Apply exponent rules

64 (4^{x})^{2} - 20 \cdot 4^{x} + 1 = 0

Rewrite the equation with

4^{x} = u

64 (u)^{2} - 20 u + 1 = 0

Solve

64 u^{2} - 20 u + 1 = 0 : u = \frac{1}{4}, u = \frac{1}{16}

u = \frac{1}{4}, u = \frac{1}{16}

Substitute back

u = 4^{x},

solve for

x

Solve

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

Solve

4^{x} = \frac{1}{16} : x = - 2

x = - 1, x = - 2

3 \cdot 9^{x} - 10 \cdot 3^{x} + 3 = 0

6^{x} = 23

x, 0 = e^{x y}

1 = 4^{- x}

2^{x} - 2^{- x} = 20

What is the answer to 64(4^{2x})-20(4^x)+1=0 ?
The answer to 64(4^{2x})-20(4^x)+1=0 is x=-1,x=-2
Solve for x: 64(4^{2x})-20(4^x)+1=0
The solution is x=-1,x=-2