What is the answer to 32(2^{2x})-65(2^x)+2=0 ?

The answer to 32(2^{2x})-65(2^x)+2=0 is x=1,x=-5

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32(2^{2x})-65(2^x)+2=0

32 (2^{2 x}) - 65 (2^{x}) + 2 = 0

x = 1, x = - 5

Solution steps

32 (2^{2 x}) - 65 (2^{x}) + 2 = 0

Apply exponent rules

32 (2^{x})^{2} - 65 \cdot 2^{x} + 2 = 0

Rewrite the equation with

2^{x} = u

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

Solve

32 u^{2} - 65 u + 2 = 0 : u = 2, u = \frac{1}{32}

u = 2, u = \frac{1}{32}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 2 : x = 1

Solve

2^{x} = \frac{1}{32} : x = - 5

x = 1, x = - 5

x, y^{-} = ln (e^{2 x} - 1)

3^{x^{2} - x - 12} = 7^{x^{2} - x - 12}

e^{6 x} - e^{3 x} - 2 = 0

4^{x - 1} = 5^{x + 1}

9^{8 x - 4} = \frac{1}{( 81 ) ^{2 x - 5}}

What is the answer to 32(2^{2x})-65(2^x)+2=0 ?
The answer to 32(2^{2x})-65(2^x)+2=0 is x=1,x=-5
Solve for x: 32(2^{2x})-65(2^x)+2=0
The solution is x=1,x=-5