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

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

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2^{2x}+64=20(2^x)

2^{2 x} + 64 = 20 (2^{x})

x = 4, x = 2

Solution steps

2^{2 x} + 64 = 20 (2^{x})

Apply exponent rules

(2^{x})^{2} + 64 = 20 \cdot 2^{x}

Rewrite the equation with

2^{x} = u

(u)^{2} + 64 = 20 u

Solve

u^{2} + 64 = 20 u : u = 16, u = 4

u = 16, u = 4

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 16 : x = 4

Solve

2^{x} = 4 : x = 2

x = 4, x = 2

4^{x + 1} = 2

8 (2^{2 x}) - 129 (2^{x}) + 16 = 0

2^{2 b - 3} = 32

2^{x - 4} = 7^{x}

3^{x + 2} = 5^{2 x + 7}

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