Proof the logaritmic identity Please help me proving the basic logarithmic identity $\log_3 12=1+\log_5 4\cdot \log_6 5\cdot \log_3 6$
 A: 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) $\log_a a=1$
$\log_3 12=1+\log_5 4\cdot \log_6 5\cdot \log_3 6$
$\log_3 12=1+\frac{\log 4}{\log 5}\cdot \frac{\log 5}{\log 6}\cdot \frac{\log 6}{\log 3}$
$\log_3 12=1+\frac{\log 4}{\log 3}$
$\log_3 12=\log_3 3+\log_3 4$
$\log_3 12=\log_3 12$
A: $1+\log_54\cdot \log_65\cdot \log_36$
$=1+\frac{\log4\cdot \log 5\cdot \log6}{\log 5\cdot \log 6 \cdot \log 3}$
$=1+\frac{\log4}{\log3}=\frac{\log3}{\log3}+\frac{\log4}{\log3}$
$=\log_33+\log_34=\log_3(3.4)=\log_312$
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.
A: 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.
A: Let $\gcd(a,b) = 1$ and $x = a^nb$ where $a,b,n \in \mathbb{Z}$, then 
$$\begin{aligned}
\log_a x &= n + \frac{\ln b}{\ln a} = n + \frac{\ln b}{\ln(b+1)}\frac{\ln(b+1)}{\ln(b+2)}\cdots \frac{\ln(b+(m-1))}{\ln(b+m)}\frac{\ln (b+m)}{\ln a} \newline 
&= n + \log_{b+1} b\log_{b+2}{b+1}\cdots \log_{b+m} (b + (m-1)) \ln_{a} (b+m).
\end{aligned}$$
Take case that $a = 3$ and $b = 4$, $n = 1$ and $m = 2$, then $\log_3 12 = 1 + \log_5 4 \log_6 5 \log_3 6$.
