Prove that $0<\frac{\ln(\frac{1}{\ln2})}{\ln2}<1$ How can I prove this statement?
$$0<\frac{\ln(\frac{1}{\ln2})}{\ln2}<1$$
I have tried using the fact that $1=\frac{1}{\ln e}<\frac{1}{\ln 2}$.
I have also tried writing it as $\log_2(\frac{1}{\ln2})$, but with no luck.
I need to prove that to show that a certain function is negative.
Any help?
Thanks!
 A: $$e<2^2<e^2$$
$$\implies 1<2\ln(2)<2$$
$$\implies \frac 12<\ln(2)<1$$
$$\implies 1<\frac{1}{\ln(2)}<2$$
$$\implies 0<\ln(\frac{1}{\ln(2)})<\ln(2)$$
$$\implies 0<\frac{\ln\left(\frac{1}{\ln(2)}\right)}{\ln(2)}<1$$
A: The statement is equivalent to showing that $$\ln 2>\frac{1}{2}\iff2\ln 2>1\iff\ln4>1.$$
Since this is clearly true, since $2<e<3$, we now know that $$\ln 2>\frac{1}{2}.$$
This means that $$\frac{1}{\ln 2}<2$$ and therefore
$$\ln\left(\frac{1}{\ln 2}\right)<\ln 2$$
and therefore
$$\frac{\ln \frac{1}{\ln 2}}{\ln 2}<1.$$
The fact that the expression is positive is quite obvious, since it is equal to $2$ positive quantities divided by one another, which is always positive.

I hope that helps. If you have any questions please don't hesitate to ask :)
A: During the proof, we will use the fact that: $$ x<y \iff \ln x<\ln y $$

Blockquote

Let's prove that the fraction is positive:
$$  2<e \iff \ln 2 < \ln e = 1 \iff \frac{1}{\ln 2} > 1 \iff \\ \iff \ln \left(\frac{1}{\ln 2} \right) > \ln 1 =0 \ \ \textrm{Numerator is positive.}  $$
$$   2>1 \iff \ln 2 > \ln 1 = 0 \ \ \textrm{Denominator is positive. }$$
Both the numerator and denominator are positive. Then the fraction must be positive.
Let's prove that the fraction is smaller than 1:
$$e < 4 \iff \sqrt e < 2 \iff \ln e^{1/2} < \ln 2 \iff \frac{1}{2}< \ln 2 \iff \frac{1}{\ln 2} < 2 \iff \\ \iff \ln\left( \frac{1}{\ln 2}\right) < \ln 2  \iff \frac{\ln\left( \frac{1}{\ln 2}\right)}{\ln 2 } < 1$$
Then $ 0<\frac{\ln\left( \frac{1}{\ln 2}\right)}{\ln 2 }<1  $.  $\\ \blacksquare$
