Question

If $x=lo{g}_{5}3+lo{g}_{7}5+lo{g}_{9}7$, then x $\ge$

• A. $\frac{3}{2}$
• B. $\frac{1}{2^{\frac{1}{3}}}$
• C. $\frac{3}{2^{\frac{1}{3}}}$
• D. None

Answer 1

The correct option is C $\frac{3}{2^{\frac{1}{3}}}$

Apply $A.M.\ge G.M.$
$\therefore x\ge 3.(lo{g}_{5}3.lo{g}_{7}5.lo{g}_{9}7{)}^{\frac{1}{3}}$
$=3(lo{g}_{9}3{)}^{\frac{1}{3}}=3(lo{g}_{3^2}3{)}^{\frac{1}{3}}=3{\left(\frac{1}{2}\right)}^{\frac{1}{3}}$