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

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

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

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

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

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

Decimal Notation

x = 1.19126 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

36 e^{5 - x} - 6^{2 x} = 0

1 0^{x} \cdot 1 0^{x + 2} = 1000

e^{5 x} = 17

\frac{3}{2} e^{x} - e^{- x} = 0

2^{(x - 2) (2 x + 1)} = 1

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