# need to show abs(ln(x)) <= 1/sqrt(x) on (0,1]

• Feb 28th 2009, 09:57 PM
lllll
need to show abs(ln(x)) <= 1/sqrt(x) on (0,1]
I'm having a great of difficulty showing $\displaystyle |\ln(x)| \leq \frac{1}{\sqrt{x}}$ on the interval $\displaystyle (0,1]$

I was thinking that $\displaystyle |\ln(x)|$ can be written as $\displaystyle -\ln(x)$ since the given values will simply be reflected, which would give me

$\displaystyle 0 \leq \underbrace{\frac{1}{\sqrt{x}}}_{\mbox{always}\ >0}+\ln(x)$

but I'm stuck with the natural logarithm.

another think that I was thinking about was to show that the function never cross on the given interval, but I don't know how
that would be so since the both tend to infinity as they approach 0.
• Mar 1st 2009, 03:57 AM
kalagota
Quote:

Originally Posted by lllll
I'm having a great of difficulty showing $\displaystyle |\ln(x)| \leq \frac{1}{\sqrt{x}}$ on the interval $\displaystyle (0,1]$

I was thinking that $\displaystyle |\ln(x)|$ can be written as $\displaystyle -\ln(x)$ since the given values will simply be reflected, which would give me

$\displaystyle 0 \leq \underbrace{\frac{1}{\sqrt{x}}}_{\mbox{always}\ >0}+\ln(x)$

but I'm stuck with the natural logarithm.

another think that I was thinking about was to show that the function never cross on the given interval, but I don't know how
that would be so since the both tend to infinity as they approach 0.

hmm, you can use MVT on $\displaystyle f(x)=\frac{1}{\sqrt{x}}+\ln(x)$ on the interval $\displaystyle (x,1), x>0$.
• Mar 1st 2009, 07:45 AM
Moo
Hello,

Consider $\displaystyle f(x)=\frac{1}{\sqrt{x}}+\ln(x)$

Compute its derivative, and set it =0.
You'll get the equation $\displaystyle x-\frac{\sqrt{x}}{2}=0$, after multiplying the equation by $\displaystyle x^2$ (which is possible since $\displaystyle x \neq 0$)
and this will give only one solution, $\displaystyle \alpha$

note that $\displaystyle f(\alpha)>0$
now is it a minimum ?
$\displaystyle \lim_{x \to 1} f(x)=1> f(\alpha)$
$\displaystyle f(1/2)=\sqrt{2}-\ln(2) \approx 0.721 > f(\alpha)$