How can we solve: \(\displaystyle\sqrt{{{x}}}+{\ln{{\left({x}\right)}}}-{1}={0}?\)

Kason Palmer

Kason Palmer

Answered question

2022-03-26

How can we solve:
x+ln(x)1=0?

Answer & Explanation

zalutaloj9a0f

zalutaloj9a0f

Beginner2022-03-27Added 17 answers

Both x and lnx are increasing functions of x, so x+lnx=1 can have at most one solution. As you note, it does have one, namely x=1, but that must be the only one: x+lnx<1 when 0<x<1, and x+lnx>1 when x>1.
Marcos Boyer

Marcos Boyer

Beginner2022-03-28Added 12 answers

x+ln(x)1=0
x=eu
eu2+u1=0
1u=eu2
(1u)eu2=1
(1u)2eu2e12=e122
(1u)2e(1u)2=e122
(1u)2=p
pep=12e12
We can see easily that p=12
(1u)2=12
u=0
x=eu=e0=1
Sometimes we cannot find p easily. If we continue to solve general way for such equations from pep=m
p=W(m)
where W(x) is Lambert W function
(1u)2=W(e122)
u=12W(e122)
x=eu=e12W(e2)=1

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