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

The answer to 3^{1-2x}=4x is x=(\W_{0}((3ln(3))/2))/(2ln(3))

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

The solution is x=(\W_{0}((3ln(3))/2))/(2ln(3))

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3^{1 - 2 x} = 4 x

x = \frac{W _{0} ( \frac{3 l n ( 3 )}{2} )}{2 ln ( 3 )}

Decimal

x = 0.34863 \dots

Solution steps

3^{1 - 2 x} = 4 x

Prepare

3^{1 - 2 x} = 4 x

for Lambert form:

1 = 4 x e^{l n (3) (- 1 + 2 x)}

Rewrite the equation with

2 x ln (3) = u

and

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

1 = 4 (\frac{u}{2 ln ( 3 )}) e^{l n (3) (- 1 + 2 (\frac{u}{2 l n ( 3 )}))}

Simplify

1 = \frac{2 e ^{- l n (3) + u} u}{ln ( 3 )}

Rewrite

1 = \frac{2 e ^{- l n (3) + u} u}{ln ( 3 )}

in Lambert form:

e^{u} u = \frac{3 ln ( 3 )}{2}

Solve

e^{u} u = \frac{3 ln ( 3 )}{2} : u = W_{0} (\frac{3 ln ( 3 )}{2})

u = W_{0} (\frac{3 ln ( 3 )}{2})

Substitute back

u = 2 x ln (3),

solve for

x

Solve

2 x ln (3) = W_{0} (\frac{3 ln ( 3 )}{2}) : x = \frac{W _{0} ( \frac{3 l n ( 3 )}{2} )}{2 ln ( 3 )}

x = \frac{W _{0} ( \frac{3 l n ( 3 )}{2} )}{2 ln ( 3 )}

(\frac{1}{25})^{2 x + 3} = 12 5^{9 x + 8}

5^{t} = 1637

1 6^{2 x + 1} = 4^{3 x + 3}

y, e^{x + y} = 0

3^{l o g_{2} (8) - x} = 9

What is the answer to 3^{1-2x}=4x ?
The answer to 3^{1-2x}=4x is x=(\W_{0}((3ln(3))/2))/(2ln(3))
Solve for x: 3^{1-2x}=4x
The solution is x=(\W_{0}((3ln(3))/2))/(2ln(3))