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How should I prove that $\log_{2}{3} < \log_{3}{6}$?

I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothing.

mathlove
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Iuli
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    Interestingly, the function $$\log_x(x+1)-1-\log_{x+1}x$$ is positive on $(1,x_)$ and negative on $(x_,\infty)$, where $$x_*\approx 1.95252$$ Hence, indeed the result asked (which is the case $x=2$) holds but proofs that it holds should be able to discriminate between the cases $x=2$ and, say, $x=1.95$... Which is definitely not encouraging. – Did Nov 01 '16 at 10:29
  • Likewise, $$\log_2x=\log_x(2x)$$ has a root at $$x^*\approx3.0696$$ which is suspiciously close to the value of interest $x=3$. – Did Nov 01 '16 at 10:32

7 Answers7

8

$\log_23$ vs $\log_36$

Multiply each term by $5$:

$5\log_23$ vs $5\log_36$

Apply logarithm rules:

$\log_23^5$ vs $\log_36^5$

Simplify:

$\log_2243$ vs $\log_37776$

Conclude:

$\log_2243<\log_2256=8=\log_36561<\log_37776$

Hence $\log_23<\log_36$

barak manos
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  • Brilliant, you just calculated $log_2 3$ and $log_3 6$ and see which one is bigger :D – Djura Marinkov Nov 01 '16 at 10:51
  • @DjuraMarinkov: Well, not quite. But in order to conclude the direction of the inequality, I applied an identical sequence of "direction-preserving" operations on each term ("preserving" - in opposed to multiplying by a negative number, for example). Anyhow... Thank you :) – barak manos Nov 01 '16 at 10:58
  • But still better then other answers – Djura Marinkov Nov 01 '16 at 11:01
  • @DjuraMarinkov: Thanks :) – barak manos Nov 01 '16 at 11:01
  • I guess it's the only way. I was trying with some inequalities (arithmetic, geometric...) but in the end it would be also an approximation – Djura Marinkov Nov 01 '16 at 11:47
  • Note that the idea to get this is to hammer $log_2(3)$ and $log_3(6)$ into a calculator, and see that one is slightly less, the other slightly more than $1.6 = 8/5$. Now multiply through by $5$ etc. – Torsten Schoeneberg May 30 '25 at 15:23
6

From what you did, we want to prove that $$\frac{1}{1+\log_32}\lt \log_32,$$ i.e. $$\log_32\gt \frac{\sqrt 5-1}{2},$$ i.e. $$2\gt 3^{(\sqrt 5-1)/2},$$ i.e. $$2^2\gt 3^{\sqrt 5-1},$$ i.e. $$3\cdot 2^2\gt 3^{\sqrt 5}$$ It is sufficient to prove that $$12\gt 3^{2.25},$$ i.e. $$12^2\gt 3^4\sqrt 3,$$ i.e. $$16\gt 9\sqrt 3$$ i.e. $$16^2\gt 81\cdot 3$$ which holds.

mathlove
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6

Here is mine...

$$\log_2 3=\frac{\log_2 243}5<\frac{\log_2 256}5=8/5$$ $$\log_3 2=\frac{\log_3 32}5>\frac{\log_3 27}5=3/5$$ $$\log_3 6=\log_3 2+\log_3 3>1+3/5=8/5>\log_2 3$$

Alex Silva
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Djura Marinkov
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2

Use:

  1. $$\log_2(3)=\frac{\ln(3)}{\ln(2)}$$
  2. $$\log_3(6)=\frac{\ln(6)}{\ln(3)}=\frac{\ln(2\cdot3)}{\ln(3)}=\frac{\ln(2)+\ln(3)}{\ln(3)}=1+\frac{\ln(2)}{\ln(3)}=1+\frac{1}{\log_2(3)}$$

So, you need to prove that:

$$\log_2(3)<\log_3(6)\space\space\space\Longleftrightarrow\space\space\space\log_2(3)<1+\frac{1}{\log_2(3)}$$

Jan Eerland
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2

Hint

You can use the fact that

$$\log_a(b)=\frac{\ln(b)}{\ln(a)}$$

so you can rewrite your inequation

$$\frac{\ln(3)}{\ln(2)}<\frac{\ln(6)}{\ln(3)}$$

if and only if

$$\ln^2(3)<\ln(2\times 3)\ln 2=\ln(2)(\ln(2)+\ln(3)).$$

E. Joseph
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  • See the comments on the other answer, which fully apply to the present one. (In the present case, additionally, the mention "Hint" can be viewed as misleading.) – Did Nov 01 '16 at 10:19
  • Why is it true? – Iuli Nov 01 '16 at 10:19
2

This is essentially the same answer as barak manos's and Djura Marinkov's, mostly just presented in a different fashion, as one long string of self-explanatory equalities and inequalities:

$$\begin{align} 5\log_23&=\log_2243\\ &\lt\log_2256\\ &=8\\ &=5+\log_327\\ &\lt5+\log_332\\ &=5\log_33+5\log_32\\ &=5\log_36 \end{align}$$

Barry Cipra
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1

We seek a solution using continued fractions. Everything is done here using only rational operations.

We begin by setting bounds on a logarithm $\log_a(b)$, with $b>a$, that will be used in the continuef fraction calculations. From the Binomial Theorem we can prive

$\forall (x>-1,y>1), (1+x)^y=1+xy+x^2\binom{y}{2}+x^3\binom{y}{3}+...>1+xy.$

For negative $x$ where the terms alternate signs, we use the condition in the $\forall$ hypothesis to show that all successive pairs of terms have positive sums.

If we then render $x=a-1,y=\log_a(b)$ with $b>a$, this lemma leads to the bound

$\log_a(b)<\dfrac{b-1}{a-1}.$

We may also try $x=(1/a)-1,y=\log_a(b)$ to obtain

$\log_a(b)>\dfrac{a}{b}×\dfrac{b-1}{a-1}.$

So

$\dfrac{a}{b}×\dfrac{b-1}{a-1}\log_a(b)<\dfrac{b-1}{a-1}.\tag{1}$

We now use this bound to calculate successive continued fraction entries for $\log_23$. Thus putting $a=2,b=3$ (1) gives

$(4/3)<\log_23<2$,

from which we identify the integer part as $1$ and compute

$\log_23=1+\log_2(3/2)=[1;\log_{3/2}2].$

We then apply (1) to $\log_{3/2}2$ obtaining

$(3/2)<\log_{3/2}2<2,$

which again identifies the integer part as $1$ so

$\log_{3/2}2=1+\log_{3/2}(4/3)=[1;\log_{4/3}(3/2)]$ and thus

$\log_23=[1;1,\log_{4/3}(3/2)].$

Using Microsoft Excel to assist the computations but only the rational operations identified here, we may obtain several terms for the continued fraction:

$\log_23=[1;1,1,2,2,3,1,...].$

We can calculate the continued fractiin terms for $\log_36$, but a shortcut is to observe that the first iteration gives

$\log_36=[1;\log_23]$

and we have the continued fraction expansion for $\log_23$ already. Thus

$\log_36=[1;1,1,1,2,2,3,...].$

Now to compare them. The two expansions are identical through three terms but differ in the fourth term, where $\log_23$ gives $2$ and $\log_36$ gives $1$. When an odd number of terms are matched the continued fractiin is monotonically decreasing in the next, even-numbered term(it would ne monotonically increasing in the reverse case), so $2>1$ in the fourth position implies $\log_23<\log_36$.

Oscar Lanzi
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