What is the answer to 2ln|x|=2x^{-1} ?

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

Solve for x: 2ln|x|=2x^{-1}

The solution is x= 1/(\W_{0)(1)}

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2 ln ∣ x ∣ = 2 x^{- 1}

x = \frac{1}{W _{0} ( 1 )}

Solution steps

2 ln ∣ x ∣ = 2 x^{- 1}

Simplify

2 x^{- 1} : \frac{2}{x}

2 ln ∣ x ∣ = \frac{2}{x}

Divide both sides by

2

ln ∣ x ∣ = \frac{1}{x}

f (x) = g (x),

then

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

e^{l n ∣ x ∣} = e^{\frac{1}{x}}

Simplify

e^{l n ∣ x ∣} : ∣ x ∣

∣ x ∣ = e^{\frac{1}{x}}

Solve

∣ x ∣ = e^{\frac{1}{x}} : x = \frac{1}{W _{0} ( 1 )}

x = \frac{1}{W _{0} ( 1 )}

Verify Solutions:

x = \frac{1}{W _{0} ( 1 )}

True

The solution is

x = \frac{1}{W _{0} ( 1 )}

lo g_{1} (\frac{1}{x}) = 2

lo g_{10} (x + 1) = lo g_{10} (2 x - 5)

x, 3.5 = - lo g_{10} (x)

lo g_{1} (2 x) = 3

1 - ln (2 x - 3) = 0

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