What is the answer to (2x-1)+ln(x)=0 ?

The answer to (2x-1)+ln(x)=0 is x=(\W_{0}(2e))/2

Solve for x: (2x-1)+ln(x)=0

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

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(2 x - 1) + ln (x) = 0

x = \frac{W _{0} ( 2 e )}{2}

Solution steps

(2 x - 1) + ln (x) = 0

Move

2 x

to the right side

- 1 + ln (x) = - 2 x

Move

1

to the right side

ln (x) = - 2 x + 1

f (x) = g (x),

then

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

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

Simplify

e^{l n (x)} : x

x = e^{- 2 x + 1}

Solve

x = e^{- 2 x + 1} : x = \frac{W _{0} ( 2 e )}{2}

x = \frac{W _{0} ( 2 e )}{2}

Verify Solutions:

x = \frac{W _{0} ( 2 e )}{2}

True

The solution is

x = \frac{W _{0} ( 2 e )}{2}

2 ln (x) = - 8

lo g_{e} (\frac{x}{6}) = a

lo g_{3} (n^{2} - 3 n + 5) = 2

0.38 = ln (x)

lo g_{2} (3 x + 1) = lo g_{2} (2 x)

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