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 $\log_23.$
Note that$$\frac{\sqrt{5}+1}{2}>\log_23\Longleftrightarrow2^{\sqrt5}>\frac92$$
we have
$$2^{\sqrt5}>2^{\sqrt{\frac{121}{25}}}=2^{11/5}>\frac92$$
This is true, because
$$2^{11/5}>\frac92 \Longleftrightarrow 2^{11}>\left(\frac92\right)^5\Longleftrightarrow2^{16}>9^5\Longleftrightarrow 2^8>3^5\Longleftrightarrow256>243$$
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
$\ln(x)=2\sum_{m=1}^\infty (\frac{x-1}{x+1})^{2m-1}/(2m-1)$
Then
$\ln 2>2×[\frac{2-1}{2+1}+\frac13(\frac{2-1}{2+1})^3]=56/81$
$\ln 3<2×(\frac{3-1}{3+1})+\color{blue}{2\sum_{m=2}^\infty (\frac{x-1}{x+1})^{2m-1}/(3)}=1+1/9 = 10/9$
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
$\log_2(3)<(10/9)/(56/81)=45/28.$
($\log_2(3)\approx1.585, 45/28\approx 1.607$)
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:
$\log_2(3)=1+\log_2(3/2)$
$\log_{3/2}2=1+\log_{3/2}(4/3)$
$\log_{4/3}(3/2)=1+\log_{4/3}(9/8)$
$\log_{9/8}(4/3)=\color{blue}{2}+\log_{9/8}(256/243)$
(cf. Ryzard Szwarc's answer)
$[1,1,1,2,...]<[1,1,1,1,...]$
$\log_2(3)(\approx 1.585)<\frac{1+\sqrt5}{2}(\approx 1.618).$
As a bonus, this method renders $[1,1,1,2]=8/5=1.6$ as the lowest terms fraction lying between the given numbers.
As an alternative, by golden ratio properties with $\phi=\frac{\sqrt{5}+1}{2}$ we have
$$\phi^2-\phi-1=0$$
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: $\log_{2}{3} < \log_{3}{6}$.
