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

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

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

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

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3^{1 - (2 \cdot 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 )}

e^{2} x - 3 e^{x} + 2 = 0

e^{z} = 2

e^{z} = 1

2^{- 2 x - 5} = 6^{4 - 4 x}

n^{n} = 6

What is the answer to 3^{1-(2*x)}=4^x ?
The answer to 3^{1-(2*x)}=4^x is x=(ln(3))/(2ln(6))
Solve for x: 3^{1-(2*x)}=4^x
The solution is x=(ln(3))/(2ln(6))