What is the answer to 2^{2x+1}=5 ?

The answer to 2^{2x+1}=5 is x=(ln(5)-ln(2))/(2ln(2))

Solve for x: 2^{2x+1}=5

The solution is x=(ln(5)-ln(2))/(2ln(2))

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2^{2 x + 1} = 5

x = \frac{ln ( 5 ) - ln ( 2 )}{2 ln ( 2 )}

Decimal Notation

x = 0.66096 \dots

Solution steps

2^{2 x + 1} = 5

Apply exponent rules

(2 x + 1) ln (2) = ln (5)

Solve

(2 x + 1) ln (2) = ln (5) : x = \frac{ln ( 5 ) - ln ( 2 )}{2 ln ( 2 )}

x = \frac{ln ( 5 ) - ln ( 2 )}{2 ln ( 2 )}

2^{2 x + 1} = 0

3 \cdot 9^{x} + 8 \cdot 3^{x} - 3 = 0

12 5^{x} = \frac{1}{5}

7 (3^{x}) = 2 (4^{(5 x - 4)})

e^{x} - 6 x = 0

What is the answer to 2^{2x+1}=5 ?
The answer to 2^{2x+1}=5 is x=(ln(5)-ln(2))/(2ln(2))
Solve for x: 2^{2x+1}=5
The solution is x=(ln(5)-ln(2))/(2ln(2))