Compare $\log_5(\log_4 3)$ and $\log_6(\log_6 3)$

Question

Let $a=\log_5(\log_4 3)$ and $b=\log_6(\log_6 3)$.
How to compare the two numbers if numerical solution is not used?

The following is my thoughts, but it did not work.

From the question, I get $5^a=\log_4 3$ and $6^b=\log_6 3$.

As $1 >\log_4 3 > \log_6 3>0$, I get $a < 0, b < 0$ and $5^a > 6^b$.

Then it holds $a > b \log_5 6$.

But $b \log_5 6 > b$ does not hold as $b<0$. I can not go on with it. Can someone give any new ideas?

Answer 1

My solution might be a little complicated.

$$ \begin{aligned} f(x)&=\log_5(\log_4x)-\log_6(\log_6x)\\ &=\frac{\ln\dfrac{\ln x}{\ln4}}{\ln5}-\frac{\ln\dfrac{\ln x}{\ln6}}{\ln6}\\ &=\ln(\ln x)\left(\frac{1}{\ln5}-\frac{1}{\ln6}\right)+\left[\frac{\ln(\ln6)}{\ln6}-\frac{\ln(\ln4)}{\ln5}\right] \end{aligned} $$ is monotonically increasing, so $f(3)>f(\mathrm{e})=\dfrac{\ln(\ln6)}{\ln6}-\dfrac{\ln(\ln4)}{\ln5}$.
$$ \begin{aligned} g(x)&=\frac{\ln(\ln x)}{\ln x}\\ g'(x)&=\frac{\ln x\dfrac{1}{x\ln x}-\ln(\ln x)\dfrac{1}{x}}{(\ln x)^2}\\ &=\frac{1-\ln(\ln x)}{x(\ln x)^2} \end{aligned} $$ When $x<\mathrm{e}^{\mathrm{e}}$, $g'(x)>0$ so $g(x)$ is monotonically increasing and $g(4)<g(6)$, and thus $$ \frac{\ln(\ln6)}{\ln6}>\frac{\ln(\ln4)}{\ln4}>\frac{\ln(\ln4)}{\ln5} $$ $f(3)>f(\mathrm{e})>0$.

Answer 2

We provide an elementary proof for the readers.

---

Observe that :

$$ \begin{align}&\frac 23>\log_32>\frac 12\\ \iff &3^4>2^6>3^3\end{align} $$

Using the above inequality, we obtain the sufficient bounds :

$$ \begin{align}\log_5(\log_43)&=-\log_5(\log_34)\\ &=-\log_5\left(2\log_32\right)\\ &>-\log_5\left(2\cdot \frac 23\right)\\ &=-\log_5\left(\frac 43\right)\end{align} $$

and

$$ \begin{align}\log_6(\log_63)&=-\log_6(\log_36)\\ &=-\log_6\left(1+\log_32\right)\\ &<-\log_6\left(1+\frac 12\right)\\ &=-\log_6\left(\frac 32\right)\end{align} $$

---

Then, we need to prove :

$$\begin{align}\log_6\left(\frac 32\right)>\log_5\left(\frac 43\right)\end{align}$$

Since

$$\begin{align}\log_6\left(\frac 32\right)>\frac 15&\iff \frac {3^5}{2^5}>6\\ &\iff 243>192\end{align}$$

It suffices to show that :

$$\begin{align}5^{\frac 15}>\frac 43&\iff 5>\frac {4^5}{3^5}\\ &\iff 1215>1024\end{align}$$

which completes the proof.

Answer 3

$\log_a x = \frac {\ln x}{\ln a}$

$\log_5(\log_4 3) = \frac {\ln 3 - \ln 4}{\ln 5}$

And

$\log_6(\log_6 3) = \frac {\ln 3 - \ln 6}{\ln 6}$

if $\log_5(\log_4 3) > \log_6(\log_6 3)$

$\frac {\ln(\frac{\ln 3}{\ln 4})}{\ln 5} > \frac {\ln (\frac {\ln 3}{\ln 6})}{\ln 6}$

$\frac {\ln(\ln 3) - \ln(\ln 4)}{\ln 5} > \frac {\ln (\ln 3) - \ln(\ln 6)}{\ln 6}$

$\frac {\ln \ln 3}{\ln 5} - \frac{\ln \ln 4}{\ln 5} > \frac{\ln \ln 3}{\ln 6} - \frac {\ln \ln 6}{\ln 6}$

Comparing terms

$\frac {\ln \ln 3}{\ln 5} > \frac{\ln \ln 3}{\ln 6}$

and

$- \frac{\ln \ln 4}{\ln 5} > - \frac {\ln\ln 6}{\ln 6}$

The last line is true, but it isn't obvious.