Proof $\log_b a\cdot\log_c b\cdot\log_a c=1$, Please help me proof $\log_b a\cdot\log_c b\cdot\log_a c=1$, where $a,b,c$ positive number different for 1.
 A: Change all to the natural logarithm $\log\,$:
$$\log_ba\cdot\log_cb\cdot\log_ac=\frac{\log a}{\log b}\frac{\log b}{\log c}\frac{\log c}{\log a}$$
and voila.
A: ${\bf Hint}\quad\begin{array}{cccccc}  &\rm x^{\,I} &\rm C\quad\\
 & \ \nearrow &  \\
\rm A\!\!\!\! &         &  \downarrow \rm x^{\,J} \\  
  &  \nwarrow &  \\
  &\rm x^K           &\rm B\quad\ \ 
\end{array}\rm\  \Rightarrow\ \ IJK\, =\, 1$
A: By definition $\log_a b = \frac{\log b}{\log a}$.
A: Let $\log_b a=x\implies b^x=a,$
$ \log_c b=y\implies c^y=b$ and
$\log_a c=z\implies a^z=c$
Now, $a^z=c\implies (b^x)^z=c\implies ((c^y)^z)^x=c\implies c^{xyz}=c\implies xyz=1$ assuming $c\neq 0,1$
Thus, $xyz=1\implies \log_b a\cdot\log_c b\cdot \log_a c=1$
A: Before we prove the given identity proof this idenity
$$\log_b a\log_c b=\log_c a$$
Proof: Implement the formula $\log_a b=\frac{\log_x b}{\log_x a}$
$$\frac{\log a}{\log b}\cdot\frac{\log b}{\log c}=\frac{\log a}{\log c}=\log_c a$$
Now proof the given identity.
$$\log_b a\cdot\log_c b\cdot\log_a c=1$$
$$\log_c a\cdot\log_a c=1$$
$$\frac{1}{\log_a c}\cdot\log_a c=1$$
$$1=1$$
