Please help me proving the basic logarithmic identity $\log_3 12=1+\log_5 4\cdot \log_6 5\cdot \log_3 6$
Proof the logaritmic identity
4 Answers
Before we prove the identity mention some logaritmic identity:
1) $\log_a b=\frac{\log_x b}{\log_x a}$
2) $\log_a xy=\log_a x + \log_a y$
3) $\log_a a=1$
$\log_3 12=1+\log_5 4\cdot \log_6 5\cdot \log_3 6$
$\log_3 12=1+\frac{\log 4}{\log 5}\cdot \frac{\log 5}{\log 6}\cdot \frac{\log 6}{\log 3}$
$\log_3 12=1+\frac{\log 4}{\log 3}$
$\log_3 12=\log_3 3+\log_3 4$
$\log_3 12=\log_3 12$
$1+\log_54\cdot \log_65\cdot \log_36$ $=1+\frac{\log4\cdot \log 5\cdot \log6}{\log 5\cdot \log 6 \cdot \log 3}$ $=1+\frac{\log4}{\log3}=\frac{\log3}{\log3}+\frac{\log4}{\log3}$ $=\log_33+\log_34=\log_3(3.4)=\log_312$
as $\log_xy=\frac{\log_zy}{\log_zx}$ and $\log_zx+\log_zy=\log_zxy$ where $z$ is positive real number $\neq 1$ as $\log_1x$ is not defined.
Using the power rule: $n\log_a(x)=\log_a(x^n)$
and the product rule $\log_a(a)+\log_a(b)=\log_a(ab)$ you get:
$1+\log_5 4\cdot \log_6 5\cdot \log_3 6\\= 1+ \log_6 (5^{\log_5 4})\cdot \log_3 6$
$=1+ \log_6 (4)\cdot \log_3 6\\=1+ \cdot \log_3( 6^{\log_6 (4)})$
$=1+\log_3( 4)\\=\log_3(3)+\log_3( 4)\\=\log_3(12)$
This avoids a bit of log writing using the other proposed identity, but not that much. In the end it's a matter of taste.