Keep Logging

We will use some of the logarithm properties such as:

$\bullet \quad \log_a m^n=n\log_am\\ \bullet \quad \log_ab=\dfrac{\log_kb}{\log_ka}\\ \bullet \quad\log_a(xy)=\log_ax+\log_ay$

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Given that, $a=\log_45~,~b=log_65$

We can write them as $a=\dfrac{\ln 5}{\ln 4}~,~b=\dfrac{\ln 6}{\ln 5}$

Multiplying them, we get,

$ab=\dfrac{\ln 5}{\ln 4}\times \dfrac{\ln 6}{\ln 5}=\dfrac{\ln 6}{\ln 4}=\dfrac{\ln 3 + \ln 2}{2\ln 2}$ $\\ \implies 2ab=1+\dfrac{\ln 3}{\ln 2} \implies 2ab-1=\dfrac{\ln 3}{\ln 2}\implies \dfrac{1}{(2ab-1)}=\dfrac{\ln 2}{\ln 3}=\log_32$

$\therefore \boxed{\log_32=\dfrac{1}{(2ab-1)}}$

$\log _{ 4 }{ 5 } =a\quad and\quad \log _{ 5 }{ 6 } =b\quad \\ { 4 }^{ a }=5\quad so\quad \log _{ 5 }{ 6 } =\\ \log _{ { 4 }^{ a } }{ 6 } \Rightarrow \\ \\ \quad \frac { 1 }{ a } \log _{ 4 }{ 6 } =b\\ \\ so\quad \log _{ 4 }{ 6 } =ab\\ \log _{ 4 }{ 2.3 } =ab\\ \log _{ { 2 }^{ 2 } }{ 2 } +\log _{ { 2 }^{ 2 } }{ 3 } =ab\\ \\ \frac { 1 }{ 2 } (1+\log _{ 2 }{ 3 } )=ab\\ \log _{ 2 }{ 3 } =2ab-1 \\ \log _{ 3 }{ 2 } =\frac { 1 }{ 2ab-1 }$