What is the answer to 4(4^{2x})-17(4^x)+4=0 ?

The answer to 4(4^{2x})-17(4^x)+4=0 is x=1,x=-1

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4(4^{2x})-17(4^x)+4=0

4 (4^{2 x}) - 17 (4^{x}) + 4 = 0

x = 1, x = - 1

Solution steps

4 (4^{2 x}) - 17 (4^{x}) + 4 = 0

Apply exponent rules

(4^{x})^{2} \cdot 4^{1} - 17 \cdot 4^{x} + 4 = 0

Rewrite the equation with

4^{x} = u

(u)^{2} \cdot 4^{1} - 17 u + 4 = 0

Solve

u^{2} \cdot 4^{1} - 17 u + 4 = 0 : u = 4, u = \frac{1}{4}

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

Substitute back

u = 4^{x},

solve for

x

Solve

4^{x} = 4 : x = 1

Solve

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

x = 1, x = - 1

2^{2 x + 3} - 8 \cdot 2^{x + 1} + 8 = 0

2^{2 x - 5} = 8

7^{10} = 7^{5} \cdot 7^{f}

2^{x} - 5 = 59

4 e^{2 x} + e^{- x} = 0

What is the answer to 4(4^{2x})-17(4^x)+4=0 ?
The answer to 4(4^{2x})-17(4^x)+4=0 is x=1,x=-1
Solve for x: 4(4^{2x})-17(4^x)+4=0
The solution is x=1,x=-1