What is the answer to 3^{-2x}=2^{1-4x} ?

The answer to 3^{-2x}=2^{1-4x} is x=(ln(2))/(2(-ln(3)+2ln(2)))

Solve for x: 3^{-2x}=2^{1-4x}

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

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3^{- 2 x} = 2^{1 - 4 x}

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

Decimal Notation

x = 1.20471 \dots

Solution steps

3^{- 2 x} = 2^{1 - 4 x}

Apply exponent rules

- 2 x ln (3) = (1 - 4 x) ln (2)

Solve

- 2 x ln (3) = (1 - 4 x) ln (2) : x = \frac{ln ( 2 )}{2 ( - ln ( 3 ) + 2 ln ( 2 ) )}

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

e^{x} + x = 27

2^{- x + 2} = 6^{2 x - 3}

x, 2^{x - 2 y} = 4

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

3^{x + 4} = 2^{1 - 3 x}

What is the answer to 3^{-2x}=2^{1-4x} ?
The answer to 3^{-2x}=2^{1-4x} is x=(ln(2))/(2(-ln(3)+2ln(2)))
Solve for x: 3^{-2x}=2^{1-4x}
The solution is x=(ln(2))/(2(-ln(3)+2ln(2)))