Question

Prove that ${\log _a}n{\log _b}n + {\log _b}n{\log _c}n + {\log _c}n{\log _a}n = \dfrac{{\left( {{{\log }_a}n} \right)\left( {{{\log }_b}n} \right)\left( {{{\log }_c}n} \right)}}{{{{\log }_{abc}}n}}$.

Answer 1

Hint: First, apply the base change law on ${\log _a}n$, ${\log _b}n$ and ${\log _c}n$. Then, take out the common ${\left( {\log n} \right)^2}$ from the numerator. After that take LCM and apply product law in the numerator. Now, multiply and divide the LHS by $\log n$, so that each $\log n$ has a denominator for the base change. After that apply base change law. The result obtained is the RHS of the equation.

Formula used:
The base change law of log is,
${\log _y}x = \dfrac{{\log x}}{{\log y}}$
The product law is,
$\log xy = \log x + \log y$

Complete step-by-step answer:
${\log _a}n{\log _b}n + {\log _b}n{\log _c}n + {\log _c}n{\log _a}n$ ….. (1)
Take the LHS of the equation.
By base change law of log,
${\log _a}n = \dfrac{{\log n}}{{\log a}}$
Similarly,
$\begin{gathered}
  {\log _b}n = \dfrac{{\log n}}{{\log b}} \\
  {\log _c}n = \dfrac{{\log n}}{{\log c}} \\
\end{gathered} $
Substitute these values in equation (1),
$\dfrac{{\log n}}{{\log a}} \times \dfrac{{\log n}}{{\log b}} + \dfrac{{\log n}}{{\log b}} \times \dfrac{{\log n}}{{\log c}} + \dfrac{{\log n}}{{\log c}} \times \dfrac{{\log n}}{{\log a}}$
Multiply the terms and take out common factors from the numerator,
${\left( {\log n} \right)^2}\left( {\dfrac{1}{{\log a\log b}} + \dfrac{1}{{\log b\log c}} + \dfrac{1}{{\log c\log a}}} \right)$
Take LCM in the bracket,
${\left( {\log n} \right)^2}\left( {\dfrac{{\log c + \log a + \log b}}{{\log a\log b\log c}}} \right)$
Now, apply multiplication law in the numerator of the equation,
${\left( {\log n} \right)^2}\left( {\dfrac{{\log abc}}{{\log a\log b\log c}}} \right)$
Multiply and divide the equation by $\log n$, we get,
${\left( {\log n} \right)^2}\left( {\dfrac{{\log abc}}{{\log a\log b\log c}}} \right) \times \dfrac{{\log n}}{{\log n}}$
Tale $\log n$as the numerator for $\log a,\log b$ and $\log c$.
$\dfrac{{\log n}}{{\log a}} \times \dfrac{{\log n}}{{\log b}} \times \dfrac{{\log n}}{{\log c}} \times \dfrac{{\log abc}}{{\log n}}$
Reverse the last fraction such that $\log abc$ becomes the denominator of $\log n$,
$\dfrac{{\log n}}{{\log a}} \times \dfrac{{\log n}}{{\log b}} \times \dfrac{{\log n}}{{\log c}} \times \dfrac{1}{{\dfrac{{\log n}}{{\log abc}}}}$
Now, apply base change law in the above equation,
$\dfrac{{{{\log }_a}n \times {{\log }_b}n \times {{\log }_c}n}}{{{{\log }_{abc}}n}}$
Hence, it is proved that ${\log _a}n{\log _b}n + {\log _b}n{\log _c}n + {\log _c}n{\log _a}n = \dfrac{{\left( {{{\log }_a}n} \right)\left( {{{\log }_b}n} \right)\left( {{{\log }_c}n} \right)}}{{{{\log }_{abc}}n}}$.

Note: Logarithms are the opposite of exponentials, just as the opposite of addition is subtraction and the opposite of multiplication is division.
In other words, a logarithm is essentially an exponent that is written in a particular manner.
Logarithms can make multiplication and division of large numbers easier, because adding logarithms is the same as multiplying, and subtracting logarithms is the same as dividing.
Change of base rule law,
${\log _y}x = \dfrac{{\log x}}{{\log y}}$
Product rule law,
$\log xy = \log x + \log y$
Quotient rule law,
$\log \dfrac{x}{y} = \log x - \log y$
Power rule law,
$\log {x^y} = y\log x$