Question

# Show that ${{\log }_{b}}a{{\log }_{c}}b{{\log }_{a}}c=1.$

Hint: Use the base change property of logarithm to make each log with the same base . Change the base of each expression to either 10 or e.

We have the expression$\left( {{\log }_{b}}a \right)\left( {{\log }_{c}}b \right)\left( {{\log }_{a}}c \right)$.
For a typical calculation, we can change logarithms to base 10 or e.
Let us use the base as 10 here.
$\therefore {{\log }_{b}}a=\dfrac{{{\log }_{10}}a}{{{\log }_{10}}b}=\dfrac{\log a}{\log b}$
${{\log }_{10}}a$ can be written as $\log a.$
Similarly, \begin{align} & {{\log }_{c}}b=\dfrac{{{\log }_{10}}b}{{{\log }_{10}}c}=\dfrac{logb}{\log c} \\ & {{\log }_{a}}c=\dfrac{{{\log }_{10}}c}{{{\log }_{10}}a}=\dfrac{\log c}{\log a} \\ \end{align}
Substituting these values in the expression, we get,
$\therefore \left( {{\log }_{b}}a \right)\left( {{\log }_{c}}b \right)\left( {{\log }_{a}}c \right)=\dfrac{\log a}{\log b}\times \dfrac{\log b}{\log c}\times \dfrac{\log c}{\log a}.$
Now let us cancel out $\log a,\log b,\log c$ from the numerator and denominator we get the answer as 1.
$\left( {{\log }_{b}}a \right)\left( {{\log }_{c}}b \right)\left( {{\log }_{a}}c \right)=1.$
Hence we have proved that $\left( {{\log }_{b}}a \right)\left( {{\log }_{c}}b \right)\left( {{\log }_{a}}c \right)=1.$

Note: We solved the problem using base k as 10. We know that the base can also be taken as k = e.
Then ${{\log }_{b}}a=\dfrac{{{\log }_{e}}a}{{{\log }_{e}}b};{{\log }_{c}}b=\dfrac{{{\log }_{e}}b}{{{\log }_{e}}c};{{\log }_{a}}c=\dfrac{{{\log }_{e}}c}{{{\log }_{e}}a}.$
By multiplying these, we still get the value as 1.
$\therefore \left( {{\log }_{b}}a \right)\left( {{\log }_{c}}b \right)\left( {{\log }_{a}}c \right)=\dfrac{{{\log }_{e}}a}{{{\log }_{e}}b}\times \dfrac{{{\log }_{e}}b}{{{\log }_{e}}c}\times \dfrac{{{\log }_{e}}c}{{{\log }_{e}}a}=1.$