Question

If $x={\log }_{5}3+{\log }_{7}5+{\log }_{9}7$ then

• A. $x\ge \frac{3}{2}$
• B. $x\ge \frac{1}{\sqrt[3]{32}}$
• C. $x\ge \frac{3}{\sqrt[3]{32}}$
• D. none of these

Answer 1

The correct option is C $x\ge \frac{3}{\sqrt[3]{32}}$

We know ${\log }_{5}3,{\log }_{7}5,{\log }_{9}7$ are positive

Thus using $A.M\ge G.M$

$\frac{{\log }_{5}3+{\log }_{7}5+{\log }_{9}7}{3}\ge ({\log }_{5}3.{\log }_{7}5.{\log }_{9}7{)}^{1/3}$

$\frac{{\log }_{5}3+{\log }_{7}5+{\log }_{9}7}{3}\ge {(\frac{\log 3}{\log 5}.\frac{\log 5}{\log 7}.\frac{\log 7}{\log 9})}^{1/3}$

Using ${\log }_{a}b=\frac{\log b}{\log a}$

$\frac{{\log }_{5}3+{\log }_{7}5+{\log }_{9}7}{3}\ge (\frac{1}{2}{)}^{\frac{1}{3}}$

$\therefore x\ge \frac{3}{\sqrt[3]{32}}$