Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How could we solve $$\sqrt{x} + \ln(x) -1 = 0$$ without using Mathematica? Obviously a solution is $x = 1$, but what are the other exact solutions?

share|cite|improve this question
up vote 14 down vote accepted

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 $x>1$.

share|cite|improve this answer
Thank you. This inspired a followup where the situation is not as clear: <…; Any ideas are welcome! – bigollo Jun 6 '12 at 6:50

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


$e^{u/2} + u -1 = 0$







We can see easily that $p=1/2$




Sometimes we cannot find p easily. If we continue to solve general way for such equations from $pe^{p}=m$


where $W(x)$ is Lambert W function




Also wolframalpha verified that result

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.