# How to prove $\frac{\sqrt{5}+1}{2}>\log_23$?

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

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

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

• Nice trick! like it :)
– 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

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

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.

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

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: $\log_{2}{3} < \log_{3}{6}$.