"How can we solve:x+ln⁡(x)−1=0?" was answered by our experts

Kason Palmer

Answered question

2022-03-26

How can we solve:

\sqrt{x} + \ln (x) - 1 = 0 ?

zalutaloj9a0f

Beginner2022-03-27Added 17 answers

Both

\sqrt{x}

and

\ln x

are increasing functions of x, so

\sqrt{x} + \ln x = 1

can have at most one solution. As you note, it does have one, namely

x = 1

, but that must be the only one:

\sqrt{x} + \ln x < 1

when

0 < x < 1

, and

\sqrt{x} + \ln x > 1

when

x > 1

Marcos Boyer

Beginner2022-03-28Added 12 answers

\sqrt{x} + \ln (x) - 1 = 0

x = e^{u}

e^{\frac{u}{2}} + u - 1 = 0

1 - u = e^{\frac{u}{2}}

(1 - u) e^{- \frac{u}{2}} = 1

\frac{(1 - u)}{2} e^{- \frac{u}{2}} e^{\frac{1}{2}} = \frac{e^{\frac{1}{2}}}{2}

\frac{(1 - u)}{2} e^{\frac{(1 - u)}{2}} = \frac{e^{\frac{1}{2}}}{2}

\frac{(1 - u)}{2} = p

p e^{p} = \frac{1}{2} e^{\frac{1}{2}}

We can see easily that

p = \frac{1}{2}

\frac{(1 - u)}{2} = \frac{1}{2}

u = 0

x = e^{u} = e^{0} = 1

Sometimes we cannot find p easily. If we continue to solve general way for such equations from

p e^{p} = m

p = W (m)

where W(x) is Lambert W function

\frac{(1 - u)}{2} = W (\frac{e^{\frac{1}{2}}}{2})

u = 1 - 2 W (\frac{e^{\frac{1}{2}}}{2})

x = e^{u} = e^{1 - 2 W (\frac{\sqrt{e}}{2})} = 1

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