I want to compare the magnitude of $2\cos36^\circ$ and $\log_23$ without using calculators.

Using $\sin72^\circ=\cos18^\circ$, I can get $$4\sin18^\circ\left(1-2\sin^218^\circ\right)=1$$

then I can get $$2\cos36^\circ=\dfrac{\sqrt{5}+1}{2},$$

but I don't know how to compare the size of $\dfrac{\sqrt{5}+1}{2}$ and $\log_23.$


4 Answers 4


Note that$$\frac{\sqrt{5}+1}{2}>\log_23\Longleftrightarrow2^{\sqrt5}>\frac92$$

we have


This is true, because

$$2^{11/5}>\frac92 \Longleftrightarrow 2^{11}>\left(\frac92\right)^5\Longleftrightarrow2^{16}>9^5\Longleftrightarrow 2^8>3^5\Longleftrightarrow256>243$$

  • 1
    $\begingroup$ Nice trick! like it :) $\endgroup$
    – user1026811
    Jun 7, 2023 at 16:43

We have $\sqrt{5}\ge 2.2.$ It suffices to show that $1.6>\log_23$ or equivalently $8>5\log_23$. The latter is equivalent to $$2^8>3^5$$ which can be verified straightforward, as $2^8=256$ and $3^5=243.$


I present two methods.

Method 1

From this answer render

$\ln(x)=2\sum_{m=1}^\infty (\frac{x-1}{x+1})^{2m-1}/(2m-1)$


$\ln 2>2×[\frac{2-1}{2+1}+\frac13(\frac{2-1}{2+1})^3]=56/81$

$\ln 3<2×(\frac{3-1}{3+1})+\color{blue}{2\sum_{m=2}^\infty (\frac{x-1}{x+1})^{2m-1}/(3)}=1+1/9 = 10/9$

In the blue expression the denominators of $2m-1$ are reduced to $3$ to get an upper bound, and then the terms are summed as a geometric series.

We then have


($\log_2(3)\approx1.585, 45/28\approx 1.607$)

Now $45^2-(45×28)-28^2=-19<0$, so $45/28$ must be less than the positive root of $x^2-x-1=0$.

Method 2

The number $(1+\sqrt5)/2$ has the continued fraction $[1,1,1,1,...]$ whereas $\log_2(3)$ will have some continued fraction $[1,a,b,c,...]$ where some entry becomes greater than $1$. If an odd-position entry is the first to be greater than $1$, then $\log_2(3)>[1,1,1,1,...]=(1+\sqrt5)/2$. If an even-position entry is the first to exceed $1$, then $\log_2(3)<[1,1,1,1,...]=(1+\sqrt5)/2$.

We set out to find the continued fraction entries for $\log_2(3)$, remembering that the reciprocal of each remainder is obtained by just reversing the arguments of the logarithm function:





(cf. Ryzard Szwarc's answer)


$\log_2(3)(\approx 1.585)<\frac{1+\sqrt5}{2}(\approx 1.618).$

As a bonus, this method renders $[1,1,1,2]=8/5=1.6$ as the lowest terms fraction lying between the given numbers.

  • $\begingroup$ $2^8>3^5\implies 2^{(1+\sqrt 5)/2}>2^{8/5}>3$. $\endgroup$ Jul 7, 2023 at 2:27

As an alternative, by golden ratio properties with $\phi=\frac{\sqrt{5}+1}{2}$ we have


therefore, since $1<\log_23<2$, it suffices to show that

$$\log^2_23-\log_23-1<0 \iff \log^2_23<\log_2 6 \iff \log_2 3 <\log_3 6$$

which is true, as proved here: Prove: $\log_{2}{3} < \log_{3}{6}$.


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