# How can we solve: $\sqrt{x} + \ln(x) -1 = 0$?

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?

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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 $x>1$.

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Thank you. This inspired a followup where the situation is not as clear: <math.stackexchange.com/questions/154577/…; Any ideas are welcome! –  bigollo Jun 6 '12 at 6:50

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

$x=e^u$

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

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

$(1-u)e^{-u/2}=1$

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

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

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

$pe^{p}=\frac{1}{2}e^{1/2}$

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

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

$u=0$

$x=e^u=e^0=1$

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

$p=W(m)$

where $W(x)$ is Lambert W function

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

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

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

Also wolframalpha verified that result

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