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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.

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

7 Answers 7

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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:

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

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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 .

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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 $${\ln 6\over \ln 5}=f(5)>f(6)={\ln 7\over \ln 6}$$

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    $\begingroup$ "obviously" -- how? $\endgroup$ Apr 6, 2023 at 5:37
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    $\begingroup$ @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 ? $\endgroup$ Apr 6, 2023 at 5:44
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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 $x > 2.$

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    $\begingroup$ Noted belatedly: @lone-student's answer (+1) uses the same argument. $\endgroup$ Apr 6, 2023 at 23:02
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    $\begingroup$ 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.) $\endgroup$ Apr 12, 2023 at 5:17
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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 $$\tag{3} 0<y, yz <1, \quad 1< z <2 $$ 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.

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

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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 $x > 1$.

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