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

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

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

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

x = 1, x = - 2

Solution steps

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

Apply exponent rules

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

Rewrite the equation with

4^{x} = u

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

Solve

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

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

Substitute back

u = 4^{x},

solve for

x

Solve

4^{x} = 4 : x = 1

Solve

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

x = 1, x = - 2

4^{1 + 2 x} \cdot 3^{4 x + 1} = 432

x, 4 - 3^{x} = 0

128 = 2^{n}

\frac{( 8 ^{x} ) ^{x}}{3 2 ^{x}} = 4

28 \cdot 3^{5 x} = 252

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