Question

Suppose ${\log }_{a}b+{\log }_{b}a=c$. The smallest possible integer value of $c$ for all $a,b>1$ is.

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

$2$

${\log }_{a}b+{\log }_{b}a=c$

Let ${\log }_{a}b=x$

$x=\frac{\log b}{\log a}{\log }_{b}a=\frac{\log a}{\log b}=\frac{1}{x}$

Using A.M$\ge$G.M, we have

$\frac{x+\frac{1}{x}}{2}\ge x\times \frac{1}{x}x+\frac{1}{x}\ge 2$

$\Rightarrow {\log }_{a}b+{\log }_{b}a\ge 2$

$\therefore c\ge 2$

Hence, option $C$ is correct.