Question

If $x = {\log _5}3 + {\log _7}5 + {\log _9}7$ then
A. $x \geqslant \dfrac{3}{2}$
B. $x \geqslant \dfrac{1}{{\sqrt[3]{2}}}$
C. $x \geqslant \dfrac{3}{{\sqrt[3]{2}}}$
D. None of these

Answer 1

Hint: We will take each of the terms in sum as three different terms and then find the arithmetic mean and the geometric mean of them so that we can use the inequality that $A.M. \geqslant G.M.$

Complete step-by-step solution:
We are given that $x = {\log _5}3 + {\log _7}5 + {\log _9}7$. ……………..(1)
Let $a = {\log _5}3,b = {\log _7}5$ and $c = {\log _9}7$. ……………(2)
We know that the arithmetic mean of the numbers a, b and c is given by the following expression:-
$ \Rightarrow $Arithmetic Mean = $\dfrac{{a + b + c}}{3}$
Let us put (2) in the above expression to get:-
$ \Rightarrow $Arithmetic Mean = $\dfrac{{{{\log }_5}3 + {{\log }_7}5 + {{\log }_9}7}}{3}$
Now putting (1) in the above expression, we will then obtain:-
$ \Rightarrow $Arithmetic Mean = $\dfrac{x}{3}$ …………..(3)
We know that the geometric mean of the numbers a, b and c is given by the following expression:-
$ \Rightarrow $Geometric Mean = ${\left( {abc} \right)^{\dfrac{1}{3}}}$
Let us put (2) in the above expression to get:-
$ \Rightarrow $Geometric Mean = ${\left( {{{\log }_5}3 \times {{\log }_7}5 \times {{\log }_9}7} \right)^{\dfrac{1}{3}}}$
Simplifying the calculations on the RHS using the fact that ${\log _a}b = \dfrac{{\log b}}{{\log a}}$
$ \Rightarrow $ Geometric Mean = ${\left( {\dfrac{{\log 3}}{{\log 5}} \times \dfrac{{\log 5}}{{\log 7}} \times \dfrac{{\log 7}}{{\log 9}}} \right)^{\dfrac{1}{3}}}$
Cutting off the similar terms in the numerator and denominator of the RHS, we will obtain:-
$ \Rightarrow $ Geometric Mean = ${\left( {\dfrac{{\log 3}}{{\log 9}}} \right)^{\dfrac{1}{3}}}$
We can write it as follows:-
$ \Rightarrow $ Geometric Mean = ${\left( {\dfrac{{\log 3}}{{\log {3^2}}}} \right)^{\dfrac{1}{3}}}$
Now we will use the fact that $\log {a^b} = b\log a$ in the above expression to get the following expression:-
$ \Rightarrow $ Geometric Mean = ${\left( {\dfrac{{\log 3}}{{2\log 3}}} \right)^{\dfrac{1}{3}}}$
Cutting off the similar terms in the numerator and denominator of the RHS, we will obtain:-
$ \Rightarrow $ Geometric Mean = ${\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{3}}}$
We can write it as follows:-
$ \Rightarrow $ Geometric Mean = $\dfrac{1}{{\sqrt[3]{2}}}$ …………….(4)
Using the equations (3), (4) in the fact that $A.M. \geqslant G.M.$, we will obtain:-
$ \Rightarrow \dfrac{x}{3} \geqslant \dfrac{1}{{\sqrt[3]{2}}}$
Multiplying by 3 on both the sides, we will get:-
$ \Rightarrow x \geqslant \dfrac{3}{{\sqrt[3]{2}}}$

$\therefore $ The correct option is (C).

Note: The students must commit to the memory the following facts and formulas:-
$ \Rightarrow $Arithmetic Mean = $\dfrac{{a + b + c}}{3}$
$ \Rightarrow $Geometric Mean = ${\left( {abc} \right)^{\dfrac{1}{3}}}$
$ \Rightarrow A.M. \geqslant G.M.$
$ \Rightarrow {\log _a}b = \dfrac{{\log b}}{{\log a}}$
$ \Rightarrow \log {a^b} = b\log a$
The students must note that we did use the concept of arithmetic mean and the geometric mean because we had to find something related to inequality. If you ever see inequality, try to use the concept of arithmetic mean and geometric mean because that clearly involves it.