Proof $\log_b a\cdot\log_c b\cdot\log_a c=1$,

Question

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.

Answer 1

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.

$$\log_b a\cdot\log_c b\cdot\log_a c=1$$

$$\log_c a\cdot\log_a c=1$$

$$\frac{1}{\log_a c}\cdot\log_a c=1$$

$$1=1$$

Answer 2

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.

Answer 3

By definition $\log_a b = \frac{\log b}{\log a}$.

Answer 4

${\bf Hint}\quad\begin{array}{cccccc} &\rm x^{\,I} &\rm C\quad\\ & \ \nearrow & \\ \rm A\!\!\!\! & & \downarrow \rm x^{\,J} \\ & \nwarrow & \\ &\rm x^K &\rm B\quad\ \ \end{array}\rm\ \Rightarrow\ \ IJK\, =\, 1$

Answer 5

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, $xyz=1\implies \log_b a\cdot\log_c b\cdot \log_a c=1$