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

The answer to 2^{2x-1}=3^{x+2} is x=(ln(18))/(2ln(2)-ln(3))

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

The solution is x=(ln(18))/(2ln(2)-ln(3))

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

x = \frac{ln ( 18 )}{2 ln ( 2 ) - ln ( 3 )}

Decimal Notation

x = 10.04710 \dots

Solution steps

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

Apply exponent rules

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

Solve

(2 x - 1) ln (2) = (x + 2) ln (3) : x = \frac{ln ( 18 )}{2 ln ( 2 ) - ln ( 3 )}

x = \frac{ln ( 18 )}{2 ln ( 2 ) - ln ( 3 )}

x, f = (\frac{1}{2})^{x}

8^{x^{2} - x} = 4^{x + 1}

e^{2 x} = \frac{1}{8}

m e^{m x} + 2 e^{m x} = 0

(100)^{x} = 0

What is the answer to 2^{2x-1}=3^{x+2} ?
The answer to 2^{2x-1}=3^{x+2} is x=(ln(18))/(2ln(2)-ln(3))
Solve for x: 2^{2x-1}=3^{x+2}
The solution is x=(ln(18))/(2ln(2)-ln(3))