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

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

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solvefor x,log_{3}(4x)=1-2x

solve for

x, lo g_{3} (4 x) = 1 - 2 x

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

Decimal Notation

x = 0.34863 \dots

Solution steps

lo g_{3} (4 x) = 1 - 2 x

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

3^{l o g_{3} (4 x)} = 3^{1 - 2 x}

Simplify

3^{l o g_{3} (4 x)} : 4 x

4 x = 3^{1 - 2 x}

Solve

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

Verify Solutions:

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

True

The solution is

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

ln (x) \cdot 2 = - ln (6)

lo g_{a} (25) = 4

4 - lo g_{10} (x^{2} - x + 4) = 3

ln (\frac{( x - 2 )}{8}) = 2

2 lo g_{10} (x + 5) + lo g_{3} (9) = 2

What is the answer to solvefor x,log_{3}(4x)=1-2x ?
The answer to solvefor x,log_{3}(4x)=1-2x is x=(\W_{0}((3ln(3))/2))/(2ln(3))