What is the answer to 4(4^{2x})-65(4^x)+16=0 ?

The answer to 4(4^{2x})-65(4^x)+16=0 is x=2,x=-1

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4(4^{2x})-65(4^x)+16=0

4 (4^{2 x}) - 65 (4^{x}) + 16 = 0

x = 2, x = - 1

Solution steps

4 (4^{2 x}) - 65 (4^{x}) + 16 = 0

Apply exponent rules

(4^{x})^{2} \cdot 4^{1} - 65 \cdot 4^{x} + 16 = 0

Rewrite the equation with

4^{x} = u

(u)^{2} \cdot 4^{1} - 65 u + 16 = 0

Solve

u^{2} \cdot 4^{1} - 65 u + 16 = 0 : u = 16, u = \frac{1}{4}

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

Substitute back

u = 4^{x},

solve for

x

Solve

4^{x} = 16 : x = 2

Solve

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

x = 2, x = - 1

e^{- 2 x} = 1

e^{3 x + 4} - 8 = 0

3^{x} = 24 3^{x - 2}

e^{x - 3} + e^{3 - x} = 22

16 + 2^{x} = 2^{x + 1}

What is the answer to 4(4^{2x})-65(4^x)+16=0 ?
The answer to 4(4^{2x})-65(4^x)+16=0 is x=2,x=-1
Solve for x: 4(4^{2x})-65(4^x)+16=0
The solution is x=2,x=-1