Simplifying a function of $\ln(x)$

Question

I was asked to convert the function

$$\frac{1}{x \ln x \sqrt{(\ln x)^2-1}}$$

into a function in the expression

$$\frac{1}{x \sqrt{f(x)}}$$

for the domain $x > e$

but I can't seem to find how I can convert it into the answer:

$$\frac{1}{x \ln x \sqrt{(\ln x)^4 - (\ln x)^2}}$$

My instinct was to use the difference of two squares:

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

but I'm still stuck because that method doesn't work.

Can someone help?

Just make $ \ln x \sqrt{(\ln x)^2-1}=\sqrt{f(x)}$ which give $ (\ln x)^2(\ln x)^2-1)=f(x)$. — Piquito, Aug 16 '18 at 09:40

score 3 · Answer 1 · answered Aug 16 '18 at 09:33

3

$f(x)=({\log x})^{2}(({\log x})^{2}-1)$ will work.

answered Aug 16 '18 at 09:33

Kavi Rama Murthy

311,013

ahdahmani · Accepted Answer · 2018-08-16T12:54:34.860

3

Only a simple calcul

If $x > e$ then

\begin{align} \frac{1}{x \ln x \sqrt{(\ln x)^2-1}}&=\frac{1}{x \ln x \sqrt{ (\frac{ln x}{ln x})^2(\ln x)^2-1}}\\ &= \frac{1}{x \dfrac{\ln x}{ln x} \sqrt{ (ln x)^2(\ln x)^2-1}}\\ &=\frac{1}{x \sqrt{ (ln x)^4-(\ln x)^2}} \end{align}

edited Aug 16 '18 at 12:54

answered Aug 16 '18 at 09:37

ahdahmani

1,298

1

Damn... thanks so much! – vik1245 Aug 16 '18 at 09:45
1

@BobSmith you are welcome :) – ahdahmani Aug 16 '18 at 09:46
1

This is wrong. For $0<x<1/e$ the function takes on negative values, but your final expression just takes on positive values. – egreg Aug 16 '18 at 10:05
@The_lost, please correct your answer, to include also the case where x is smaller than 1/e. – dmtri Aug 16 '18 at 10:28
Please see the answer below of egreg. – dmtri Aug 16 '18 at 11:00
@dmtri thank you, it is ok now ? – ahdahmani Aug 16 '18 at 11:47
Nice of you!) : – dmtri Aug 16 '18 at 11:51
@dmtri thank you :) – ahdahmani Aug 16 '18 at 11:54
@The_lost Now it’s even wronger than before – egreg Aug 16 '18 at 12:43
@egreg could you explain please? – ahdahmani Aug 16 '18 at 12:45
I should have mentioned $ x > e $ – vik1245 Aug 16 '18 at 12:50

score 3 · Answer 3 · answered Aug 16 '18 at 10:04

The function is defined for \begin{cases} x>0 \\[4px] \ln x\ne0 \\[4px] (\ln x)^2-1>0 \end{cases} The third condition (together with the first) becomes $\ln x<-1$ or $\ln x>1$, that is, $0<x<1/e$ or $x>e$.

The second condition is $x\ne1$, which is implied by the third condition.

Your function can be rewritten as $$ \frac{1}{x\ln x\sqrt{(\ln x)^2-1}}= \begin{cases} -\dfrac{1}{x\sqrt{(\ln x)^4-(\ln x)^2}} & 0<x<1/e \\[6px] \dfrac{1}{x\sqrt{(\ln x)^4-(\ln x)^2}} & x>e \end{cases} $$ It would be a mistake to just bring $\ln x$ as $(\ln x)^2$ inside the square root for the case $0<x<1/e$, because in this case the value of the function is negative.

The mistake would be the same as writing $1/\sqrt{x^2}=1/x$, which only holds for $x>0$.

Simplifying a function of $\ln(x)$

3 Answers3