Question

If $a,b\in N$ then the minimum value of ${\log }_{b}a+{\log }_{a}b$ is

• A. $2$
• B. $4$
• C. $0$
• D. $12$

Answer 1

Answer: (A)

$2$

Let ${\log }_{b}a=N$ then ${\log }_{a}b=\frac{1}{N}$

Minimum value of $N+\frac{1}{N}$

$A.M\ge G.M$

i.e. $\frac{N+\frac{1}{N}}{2}\ge \sqrt{N.\frac{1}{N}}$

$\Rightarrow N+\frac{1}{N}\ge 2$