What is the answer to 2^{2x-1}=3^{x+1} ?

The answer to 2^{2x-1}=3^{x+1} is x=(ln(6))/(2ln(2)-ln(3))

Solve for x: 2^{2x-1}=3^{x+1}

The solution is x=(ln(6))/(2ln(2)-ln(3))

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2^{2 x - 1} = 3^{x + 1}

x = \frac{ln ( 6 )}{2 ln ( 2 ) - ln ( 3 )}

Decimal Notation

x = 6.22826 \dots

Solution steps

2^{2 x - 1} = 3^{x + 1}

Apply exponent rules

(2 x - 1) ln (2) = (x + 1) ln (3)

Solve

(2 x - 1) ln (2) = (x + 1) ln (3) : x = \frac{ln ( 6 )}{2 ln ( 2 ) - ln ( 3 )}

x = \frac{ln ( 6 )}{2 ln ( 2 ) - ln ( 3 )}

5^{x + 1} + 5 + 5^{x - 1} = \frac{31}{5}

\frac{6 ^{x} - 7 ( 6 ^{- x} )}{2} = 3

5^{x} = 2 5^{10 x - 8}

2^{x^{2} + 7 x} = 8^{- 4}

x, 2 = e^{x y}

What is the answer to 2^{2x-1}=3^{x+1} ?
The answer to 2^{2x-1}=3^{x+1} is x=(ln(6))/(2ln(2)-ln(3))
Solve for x: 2^{2x-1}=3^{x+1}
The solution is x=(ln(6))/(2ln(2)-ln(3))