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

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

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x-ln(1/x)=2

x - ln (\frac{1}{x}) = 2

x = W_{0} (e^{2})

Decimal Notation

x = 1.55714 \dots

Solution steps

x - ln (\frac{1}{x}) = 2

Move

x

to the right side

- ln (\frac{1}{x}) = 2 - x

Apply log rules

ln (x) = 2 - x

f (x) = g (x),

then

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

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

Simplify

e^{l n (x)} : x

x = e^{2 - x}

Solve

x = e^{2 - x} : x = W_{0} (e^{2})

x = W_{0} (e^{2})

Verify Solutions:

x = W_{0} (e^{2})

True

The solution is

x = W_{0} (e^{2})

ln (ln (9 x)) = 0

ln (3 x + 4) - ln (2 x + 8) = 0

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

lo g_{2} (x) + 1 = lo g_{4} (x)

lo g_{6} (x^{2} + 9 x) = 2