What is the answer to 2^{2x}-5*2^{2x+2}+64=0 ?

The answer to 2^{2x}-5*2^{2x+2}+64=0 is x=(ln(64/19))/(2ln(2))

Solve for x: 2^{2x}-5*2^{2x+2}+64=0

The solution is x=(ln(64/19))/(2ln(2))

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2^{2 x} - 5 \cdot 2^{2 x + 2} + 64 = 0

x = \frac{ln ( \frac{64}{19} )}{2 ln ( 2 )}

Decimal Notation

x = 0.87603 \dots

Solution steps

2^{2 x} - 5 \cdot 2^{2 x + 2} + 64 = 0

Apply exponent rules

2 x ln (2) = ln (\frac{64}{19})

Solve

2 x ln (2) = ln (\frac{64}{19}) : x = \frac{ln ( \frac{64}{19} )}{2 ln ( 2 )}

x = \frac{ln ( \frac{64}{19} )}{2 ln ( 2 )}

3 2^{2 n + 1} = 8^{4 n - 1}

0 = 5^{x - 1} - 2

5 = 100 e^{- 0.00043 t}

10 0^{x} - 1 0^{x} - 1 = 0

39 \cdot 1 0^{4 x} = 390

What is the answer to 2^{2x}-5*2^{2x+2}+64=0 ?
The answer to 2^{2x}-5*2^{2x+2}+64=0 is x=(ln(64/19))/(2ln(2))
Solve for x: 2^{2x}-5*2^{2x+2}+64=0
The solution is x=(ln(64/19))/(2ln(2))