Question

Find the value of $\left( {{{\log }_3}12} \right)\left( {{{\log }_3}72} \right) - {\log _3}\left( {192} \right){\log _3}6$

Answer 1

Hint: First reduce all the numbers in logarithm in 2’s and 3’s by factorization. Then apply the logarithm properties $\log m \times n = \log m + \log n{\text{, }}\log {m^n} = n\log m{\text{ and }}{\log _m}m = 1$ to convert the entire expression in ${\log _3}2$. Simplify the algebraic expression further to get the final answer.

Complete step-by-step answer:
According to the question, the given logarithmic expression is $\left( {{{\log }_3}12} \right)\left( {{{\log }_3}72} \right) - {\log _3}\left( {192} \right){\log _3}6$. Let its value is $k$. Then we have:
$ \Rightarrow k = \left( {{{\log }_3}12} \right)\left( {{{\log }_3}72} \right) - {\log _3}\left( {192} \right){\log _3}6$
Factoring all the numbers in logarithms, we’ll get:
$ \Rightarrow k = \left( {{{\log }_3}4 \times 3} \right)\left( {{{\log }_3}8 \times 9} \right) - {\log _3}\left( {64 \times 3} \right){\log _3}2 \times 3$
We know that $4 = {2^2},{\text{ }}8 = {2^3}{\text{, }}64 = {2^6}{\text{ and }}9 = {3^2}$. Substituting these values above, we’ll get:
$ \Rightarrow k = \left( {{{\log }_3}{2^2} \times 3} \right)\left( {{{\log }_3}{2^3} \times {3^2}} \right) - {\log _3}\left( {{2^6} \times 3} \right){\log _3}2 \times 3$
Now, we also know that according to the property of logarithm, we have:
$ \Rightarrow \log m \times n = \log m + \log n$
Using this property for the above expression, we’ll get:
$ \Rightarrow k = \left( {{{\log }_3}{2^2} + {{\log }_3}3} \right)\left( {{{\log }_3}{2^3} + {{\log }_3}{3^2}} \right) - \left( {{{\log }_3}{2^6} + {{\log }_3}3} \right)\left( {{{\log }_3}2 + {{\log }_3}3} \right)$
We have another property of logarithm which is given as:
$ \Rightarrow \log {m^n} = n\log m$
Using this property for the logarithmic expression, we’ll get:
$ \Rightarrow k = \left( {2{{\log }_3}2 + {{\log }_3}3} \right)\left( {3{{\log }_3}2 + 2{{\log }_3}3} \right) - \left( {6{{\log }_3}2 + {{\log }_3}3} \right)\left( {{{\log }_3}2 + {{\log }_3}3} \right)$
Again we will use the property:
$ \Rightarrow {\log _m}m = 1$
On using this, we will have:
$ \Rightarrow k = \left( {2{{\log }_3}2 + 1} \right)\left( {3{{\log }_3}2 + 2} \right) - \left( {6{{\log }_3}2 + 1} \right)\left( {{{\log }_3}2 + 1} \right)$
Now, to make the simplification easier, put ${\log _3}2 = x$. From this we’ll get:
$ \Rightarrow k = \left( {2x + 1} \right)\left( {3x + 2} \right) - \left( {6x + 1} \right)\left( {x + 1} \right)$
Multiplying the terms and simplifying it further, we’ll get:
$
   \Rightarrow k = 2x\left( {3x + 2} \right) + 1\left( {3x + 2} \right) - \left[ {6x\left( {x + 1} \right) + 1\left( {x + 1} \right)} \right] \\
   \Rightarrow k = 6{x^2} + 4x + 3x + 2 - \left( {6{x^2} + 6x + x + 1} \right) \\
   \Rightarrow k = 6{x^2} + 7x + 2 - 6{x^2} - 7x - 1
 $
Here, $6{x^2}$ will cancel out with $ - 6{x^2}$ and $7x$ will cancel out with $ - 7x$ So the final value is:
$ \Rightarrow k = 2 - 1 = 1$

Therefore the value of $\left( {{{\log }_3}12} \right)\left( {{{\log }_3}72} \right) - {\log _3}\left( {192} \right){\log _3}6$ is $1$.

Note: Here is the summary of all the logarithmic properties used above:
$
   \Rightarrow {\log _m}m = 1 \\
   \Rightarrow \log {m^n} = n\log m \\
   \Rightarrow \log m \times n = \log m + \log n
 $
Some other important logarithmic properties are:
$
   \Rightarrow \log \dfrac{m}{n} = \log m - \log n \\
   \Rightarrow {\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}} \\
   \Rightarrow {\log _a}b = \dfrac{1}{{{{\log }_b}a}} \\
   \Rightarrow {\log _{{a^b}}}m = \dfrac{1}{b}{\log _a}m
 $