Question

If a and b are positive real numbers other than unity, then the least value of |log_b a + log_a b|, is

Answer 1

Correct Answer - Option 3 : 2

Concept:

AM, GM, HM Formulas

If A is the arithmetic mean of numbers a and b and is given by ⇔ \({\rm{A}} = \frac{{{\rm{a\;}} + {\rm{\;b}}}}{2}\)

If G is the geometric mean of the numbers a and b and is given by ⇔ \({\rm{G}} = \sqrt {{\rm{ab}}} \)

If H is the Harmonic mean of numbers a and b and is given by ⇔ \({\rm{H}} = \frac{{2{\rm{ab}}}}{{{\rm{a}}\ +\ {\rm{b}}}}\)

Relation between AM, GM and HM

1. G2 = AH
2. AM  ≥  GM  ≥  HM

Change of base rule

\({\log _m}n = \frac{{{{\log }_a}n}}{{{{\log }_a}m}}\)

Calculation:

\(\rm \left| {{{\log }_b}a + {\rm{ lo}}{{\rm{g}}_a}b} \right| = \left| {{{\log }_b}a + {\rm{ }}\frac{1}{{{{\log }_b}a}}} \right|\)

Let, log_b a = x

Then, \(\rm \left| {{{\log }_b}a + {\rm{ }}\frac{1}{{{{\log }_b}a}}} \right| = \left| {x + {\rm{ }}\frac{1}{x}} \right|\)

We know that,

AM ≥ GM

⇒ \(\rm \frac { x \ +\ \frac 1 x }2 \geq \sqrt {x \times \frac 1 x}\)

⇒ \(\rm x + \frac 1 x \geq 2\)

⇒ \(\rm \left| x + \frac 1 x \right| \geq 2\)

So,  \(\rm \left| {{{\log }_b}a + {\rm{ }}\frac{1}{{{{\log }_b}a}}} \right| \ge 2\)