Answer

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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.

Complete step-by-step answer:

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.

\[\therefore \left( {{\log }_{b}}a \right)\left( {{\log }_{c}}b \right)\left( {{\log }_{a}}c \right)=\dfrac{{{\log }_{e}}a}{{{\log }_{e}}b}\times \dfrac{{{\log }_{e}}b}{{{\log }_{e}}c}\times \dfrac{{{\log }_{e}}c}{{{\log }_{e}}a}=1.\]

Complete step-by-step answer:

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.

\[\therefore \left( {{\log }_{b}}a \right)\left( {{\log }_{c}}b \right)\left( {{\log }_{a}}c \right)=\dfrac{{{\log }_{e}}a}{{{\log }_{e}}b}\times \dfrac{{{\log }_{e}}b}{{{\log }_{e}}c}\times \dfrac{{{\log }_{e}}c}{{{\log }_{e}}a}=1.\]

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