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

The answer to solvefor x,3^{1-2x}=4^x is x=(ln(3))/(2ln(6))

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

solve for

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

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

Decimal Notation

x = 0.30657 \dots

Solution steps

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

Apply exponent rules

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

Solve

(1 - 2 x) ln (3) = x ln (4) : x = \frac{ln ( 3 )}{2 ln ( 6 )}

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

3^{x} = \frac{1}{27}

x, e^{2 x} = 3 y^{2}

x, 5^{x + y} = 125

e^{x + 2} = 15

x, e^{3 x + 2} = 5

What is the answer to solvefor x,3^{1-2x}=4^x ?
The answer to solvefor x,3^{1-2x}=4^x is x=(ln(3))/(2ln(6))