4

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
  • 139,939
Iuli
  • 6,790
  • 2
    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

6 Answers6

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
  • 43,109
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
  • 139,939
5

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
  • 3,557
Djura Marinkov
  • 2,595
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
  • 28,671
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
  • 14,843
  • 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
1

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
  • 79,832