# Comparing $\log_2 3$ to $\log_3 5$

Having been asked to compare $\log_2(3)$ and $\log_3(5)$, this is my proof: $$\log_2(3)>\log_3(5)$$ Then one uses the rule $\log_a(b)=\frac{1}{\log_b(a)}$, so $$\log_3(2)>\frac{1}{\log_3(5)}$$ since the denomenator is biger in the fraction, hence the proof is correct. Can anyone tell me if this proof is valid?

• Did you mean ${1\over \log_3 2} > \log_3 5$? – Andrew Li Mar 6 '18 at 6:08

Since $\sqrt{2}\approx 1.4$ and $\sqrt{3}\approx 1.7$, we have:
$$2^{1.5}\approx 2\times 1.4=2.8<3$$ and $$3^{1.5}\approx 3\times 1.7=5.1>5$$
Hence $\log_2(3)>1.5>\log_3(5)$. It was very fortunate that $1.5$ happens to fall in between the two, to allow hand calculation.
Your proof is invalid. The correct way to prove this is: \begin{align} &3^2>2^3\Rightarrow 2\log_23>3\Rightarrow\log_23>\frac{3}{2}\\ &3^3>5^2\Rightarrow 3>2\log_35\Rightarrow\frac{3}{2}>\log_35 \end{align}
In order to compare $\log_2(3)=\frac{\log 3}{\log 2}$ and $\log_3(5)=\frac{\log 5}{\log 3}$ it is enough to compare $\log(3)^2=\log(2+1)^2$ and $\log(2)\log(5)=\log(1+1)\log(4+1)$. We may notice that $\log(x+1)$ is log-concave on $\mathbb{R}^+$, since $$\frac{d^2}{dx^2}\log\log(x+1)=-\frac{1+\log(x+1)}{(1+x)^2 \log^2(1+x)}< 0$$ hence it follows that $\log(2+1)^2 > \log(1+1)\log(4+1)$ and $\log_2(3)>\log_3(5)$.