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

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

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ln(x)+x=-1

ln (x) + x = - 1

x = W_{0} (\frac{1}{e})

Solution steps

ln (x) + x = - 1

Move

x

to the right side

ln (x) = - 1 - x

f (x) = g (x),

then

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

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

Simplify

e^{l n (x)} : x

x = e^{- 1 - x}

Solve

x = e^{- 1 - x} : x = W_{0} (\frac{1}{e})

x = W_{0} (\frac{1}{e})

Verify Solutions:

x = W_{0} (\frac{1}{e})

True

The solution is

x = W_{0} (\frac{1}{e})

x, ln (x^{2}) = 14

lo g_{10} (∣ x ∣) = - 4

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

2 ln (x) = - 8

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