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Original title had a typo, third term of the LHS is $\log(11)$.

Prove that $\log_8(9)+\log_9(10)+\log(11)<2\log_2(3)$

I am kind of frustrated with this simple problem. How do you prove this without using a calculator. I know that both LHS and RHS are greater than 3. On the left hand side, $\log_2(9)>\log_2(8)=3$. On the RHS, each term should be slighly greater than 1. How should I start?

GFauxPas
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James Warthington
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2 Answers2

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Lemma: For any natural number $n\ge2$:

$$\log_n(n+1) \gt \log_{n+1}(n+2)\qquad(1)$$

Proof: Check the last inequality proved on the following page: Logarithmic ineqaulities

Now transform the inequality in the following way:

$$\log_89+\log_910 +\log_{10}11<2\log_23$$

$$\log_89+\log_910 +\log_{10}11<\log_23^2$$

$$\log_89+\log_910 +\log_{10}11<\log_{2^3}3^6$$

$$\log_89+\log_910 +\log_{10}11<\log_{8}(9\cdot9^2)$$

$$\log_89+\log_910 +\log_{10}11<\log_{8}9 + \log_89^2$$

$$\log_910 +\log_{10}11<2\log_89$$

which is obviously true, because lemma (1) guarantees that:

$$\log_89 \gt \log_910 \gt \log_{10}11$$

Saša
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  • Thank you. When I am trying to solve this, I also had that inequality in mind. There was an excercise that asks you to prove it in the book "Problems in Mathematics by Dybov, Miroshin et al. But I could not have come up with $\log_2^3(3^6)$. Vey nice! – James Warthington Jun 12 '18 at 03:34
  • If tou liked an answer, try to upvote it :) BTW, very nice inequality. – Saša Jun 12 '18 at 05:16
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HINT

Recall that

$$\log_a b = \frac{\log b}{\log a}$$

then

$$\log_8(9)+\log_9(10)+\log(10)<2\log_2(3)\iff \frac{\log 9}{\log 8}+\frac{\log 10}{\log 9}+\log 11<2\frac{\log 3}{\log 2}$$

user
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  • So you choose base 10 to change the base, I chose base 2. I am sorry, the third term should be $\log(11)$, not $\log(10)$. – James Warthington Jun 11 '18 at 14:28
  • @JamesWarthington with $log$ I mean $ln$ but also $\log_2$ could be a good choice. What have you obtained? – user Jun 11 '18 at 14:30
  • Well, I have $\frac{\log_2(9)}{log_2(8)}+\frac{\log_2(10)}{log_2(9)}+\frac{\log_2(11)}{log_2(10)}$. The first term is clearly smaller than $\log_2(9)$. The second term should also be, but I don't know how to reason the third term. All in all, none of this leads to a conclusive proof. – James Warthington Jun 11 '18 at 14:33
  • $1<\frac{\log_2(9)}{3}<\frac{\log_2(10)}{\log_2(9)}<\frac{\log_2(11)}{\log_2(10)}$. That's all I can come up with for now. Each term is clearly great than 1, so the sum should be greater than 3. But then the LHS is also greater than 3 by a small magnitude. – James Warthington Jun 11 '18 at 14:53