Question

Prove that $({\log _a}n)({\log _b}n) + ({\log _b}n)({\log _c}n) + ({\log _c}n)({\log _a}n) = \dfrac{{{{\log }_n}a + {{\log }_n}b + {{\log }_n}c}}{{{{\log }_n}abc}}$

Answer 1

Use logarithmic formulas to prove.
First we have to take left hand side and try to prove the right hand side
Second, in that sum we are going to use the property for changing the base of logarithms on the left hand side, continue with that, sometimes we have to use the product of two logarithms.
Finally we get the right hand side.

Complete step-by-step answer:
It is given that $({\log _a}n)({\log _b}n) + ({\log _b}n)({\log _c}n) + ({\log _c}n)({\log _a}n)$=$\dfrac{{{{\log }_n}a + {{\log }_n}b + {{\log }_n}c}}{{{{\log }_n}abc}}$
We have to prove the left hand side is equal to the right hand side.
First we take the Left hand side,
$\Rightarrow ({\log _a}n)({\log _b}n) + ({\log _b}n)({\log _c}n) + ({\log _c}n)({\log _a}n)...\left( 1 \right)$
Let us take $({\log _a}n),({\log _b}n),({\log _c}n)$ in equation $\left( 1 \right)$
We use the property for changing the base of logarithms, that is
$({\log _a}n) = \dfrac{1}{{{{\log }_n}a}}$
$({\log _b}n) = \dfrac{1}{{{{\log }_n}b}}$
$({\log _c}n) = \dfrac{1}{{{{\log }_n}c}}$
Substitute these values in$\left( 1 \right)$, then it become
$\dfrac{1}{{{{\log }_n}a}}\dfrac{1}{{{{\log }_n}b}} + \dfrac{1}{{{{\log }_n}b}}\dfrac{1}{{{{\log }_n}c}} + \dfrac{1}{{{{\log }_n}c}}\dfrac{1}{{{{\log }_n}a}}$
Multiply the terms
$\dfrac{1}{{{{\log }_n}a \times {{\log }_n}b}} + \dfrac{1}{{{{\log }_n}b \times {{\log }_n}c}} + \dfrac{1}{{{{\log }_n}c \times {{\log }_n}a}}...\left( 2 \right)$
Here we apply product rule for logarithmic,
${\log _e}a \times {\log _e}b = {\log _e}ab$
Here $e$ is equal to $n$
So we can write it as,
${\log _n}a \times {\log _n}b = {\log _n}ab$
${\log _n}b \times {\log _n}c = {\log _n}bc$
${\log _n}c \times {\log _n}a = {\log _n}ca$
Putting the above values in $\left( 2 \right)$ and we get,
$\Rightarrow \dfrac{1}{{{{\log }_n}ab}} + \dfrac{1}{{{{\log }_n}bc}} + \dfrac{1}{{{{\log }_n}ca}}$
On multiply and divide by ${\log _n}c$ in $\dfrac{1}{{{{\log }_n}ab}}$, ${\log _n}a$ in $\dfrac{1}{{{{\log }_n}bc}}$ and ${\log _n}b$ in $\dfrac{1}{{{{\log }_n}ca}}$
We can write it as,
$\dfrac{{{{\log }_n}c}}{{{{\log }_n}ab \times {{\log }_n}c}} + \dfrac{{{{\log }_n}a}}{{{{\log }_n}a \times {{\log }_n}bc}} + \dfrac{{{{\log }_n}b}}{{{{\log }_n}ca \times {{\log }_n}b}}...\left( 3 \right)$
Again, we use product rule for logarithmic
${\log _e}a \times {\log _e}b = {\log _e}ab$
Here $e$ is equal to$n$
${\log _n}ab \times {\log _n}c = {\log _n}abc$
${\log _n}a \times {\log _n}bc = {\log _n}abc$
${\log _n}ca \times {\log _n}b = {\log _n}abc$
So we can write that equation $\left( 3 \right)$
$\Rightarrow \dfrac{{{{\log }_n}c}}{{{{\log }_n}abc}} + \dfrac{{{{\log }_n}a}}{{{{\log }_n}abc}} + \dfrac{{{{\log }_n}b}}{{{{\log }_n}abc}}$
Here the denominator is same so we can write,
$\Rightarrow \dfrac{{{{\log }_n}c + {{\log }_n}a + {{\log }_n}b}}{{{{\log }_n}abc}}$
We can write it correct in order,
$\Rightarrow \dfrac{{{{\log }_n}a + {{\log }_n}b + {{\log }_n}c}}{{{{\log }_n}abc}}....\left( 4 \right)$
Hence left hand side is equal to right hand side $(1) = (4)$
$\therefore$ $({\log _a}n)({\log _b}n) + ({\log _b}n)({\log _c}n) + ({\log _c}n)({\log _a}n)$ =$\dfrac{{{{\log }_n}a + {{\log }_n}b + {{\log }_n}c}}{{{{\log }_n}abc}}$
Hence we proved.

Note: Here, we start with the basic property of logarithm by continuing this process, just by simplifying, we get the answer.
So, for this type of question to occur, first take the left hand side and then try to prove the right hand side.
By practicing the number of times, we get an idea for how to solve the equation.
A general form for product rule for logarithmic:
The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individuals’ logarithms.
${\log _b}\left( {mn} \right) = {\log _b}\left( m \right) + {\log _b}\left( n \right)$ for $b > 0$