# How to prove that $\log_5(6)>\log_6(7)$?

I know that $$\log_5(6)>\log_6(7)$$ but I wanted to prove it without calculating the values.

After generalizing it it turned this way (for $$x>1$$):
$$\frac{\ln(x)}{\ln(x-1)}>\frac{\ln(x+1)}{\ln(x)}$$

$$\ln(x)^2>\ln(x+1)\ln(x-1)$$

and based on the fact that $$\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(x)=\dfrac{1}{x}$$ I conclude that since $$\dfrac{1}{x-1}$$ is bigger than $$\dfrac{1}{x+1}$$ then it must have changed more so for example in the first part of question $$\dfrac{1}{5}>\dfrac{1}{7}$$ so $$\ln(5)$$ to $$\ln(6)$$ rate of change is bigger than $$\ln(6)$$ to $$\ln(7)$$ and it convinced me.
But I would like a more formal proof if there is and preferably one that doesn't use derivatives.

• I think the condition should be $x > 2,$ not $x > 1.$ Apr 6, 2023 at 14:53

By change of base, you want to show $${\ln (6) \over \ln (5)}>{\ln (7) \over \ln (6)}\iff \ln (6)>\sqrt{\ln (7)\ln (5)}.$$

This follows by AM-GM inequality:

$$\ln(6)=\frac{1}{2}\ln(36)>\frac{1}{2}\ln(35)={\ln(7)+\ln(5) \over 2}>\sqrt{\ln(7)\ln(5)}$$

Since, \begin{align}&\frac 65>\frac 76\\ \implies &\log_5\frac 65>\log_5\frac 76>\log_6\frac 76\end{align}

Then you have :

\begin{align}&\log_5\left(5\cdot \frac 65\right)>\log_6\left(6\cdot \frac 76\right)\\ \iff&\log_5\frac 65>\log_6\frac 76\end{align}

This completes the proof .

For $$x>0$$ the function $$f(x)={\ln(x+1)\over \ln x}={\ln x+\ln\left (1+{1\over x}\right)\over \ln x}\\ =1 +{1\over \ln x}\cdot{\ln\left (1+{1\over x}\right)}$$ is obviously strictly decreasing as a constant plus the product of positive strictly decreasing functions. Therefore $${\ln 6\over \ln 5}=f(5)>f(6)={\ln 7\over \ln 6}$$

• "obviously" -- how? Apr 6, 2023 at 5:37
• @TorstenSchoeneberg $(\ln x)^{-1}$ is strictly decreasing. Also $x^{-1}$ is strictly decreasing so is $\ln(1+x^{-1}).$ The product of two positive decreasing function is decreasing, isn't it ? Apr 6, 2023 at 5:44

This kind of question comes up quite often. Some examples are listed here.

In particular, this answer (2012) and this comment (2018) give this simple proof: $$\log_{x-1}x = 1 + \log_{x-1}\frac{x}{x-1} > 1 + \log_{x-1}\frac{x+1}x > 1 + \log_x\frac{x+1}x = \log_x(x+1)$$ for all $$x > 2.$$

• Noted belatedly: @lone-student's answer (+1) uses the same argument. Apr 6, 2023 at 23:02
• This answer of yours seems to me more practical than using the AM-GM inequality . (because, your method also provides for quick generalization). I've wanted to edit the answer or post a new answer to generalize the argument, but it looks like you've already made the necessary generalization. Great. $\rm{+1}$ (However, I see you prefer to post a community answer.) Apr 12, 2023 at 5:17

Given $$x>1$$, we set $$\tag{1} y=1-\frac{1}{x}, \quad z=1+\frac{1}{x}$$ so that $$\tag{2} x-1=xy, \quad x+1=xz, \quad yz=1-\frac{1}{x^2}.$$ Notice that $$\tag{3} 0 Now $$\begin{eqnarray}\tag{4} \log(x-1)&=&\log(xy)=\log(x)+\log(y)\cr \log(x+1)&=&\log(xz)=\log(x)+\log(z)\cr \log(x-1)\log(x+1)&=&\log^2(x)+[\log(y)+\log(z)]\log((x)+\log(y)\log(z)\cr &=&\log^2(x)+\log(yz)\log(x)+\log(y)\log(z) \end{eqnarray}$$ Thanks to the fact that $$x>1$$ together with (3) we have $$\tag{5} \max(\log(y), \log(yz) )<0, \quad \min(\log(x), \log(z))>0$$ Combining (5) to the last equation in (4), we obtain $$\log(x-1)\log(x+1)<\log^2(x)$$ or equivalently $$\tag{6} \log_{x-1}(x)> \log_x(x+1) \quad \forall x>1$$ Setting $$x=6$$, we get the desired inequality.

We have

$$\log_56>\log_67 \iff \left(\frac 6 5\right)^{\log_56}>\frac 76$$

which is true indeed

$$\left(\frac 6 5\right)^{\log_56}>\frac 65>\frac 76$$

Edit "Derivation of the first equivalence"

We start from $$\log_56>\log_67$$ then by exponentiation both sides we obtain $$6^{\log_56}>6^{\log_67}=7$$ and dividing by $$6$$ we have $$\frac{6^{\log_56}}6>\frac 7 6$$ and since $$6 =5^{\log_5 6}$$ we obtain $$\frac{6^{\log_56}}{5^{\log_5 6}}>\frac 7 6$$ that is $$\left(\frac 6 5\right)^{\log_56}>\frac 76$$.

We have $$\frac{\ln 6}{\ln 5}>\frac{\ln 7}{\ln 6}\Longleftrightarrow \ln 6 > \sqrt{\ln 7\ln5} \Longleftrightarrow \ln \ln 6 > \frac{\ln\ln 7 + \ln\ln 5}{2}$$ which follows because $$\ln \ln x$$ is a strictly concave function for $$x > 1$$.