Suppose log _a b + log _b a = c. The smallest | vedclass

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(C) Given $\log _a b + \log _b a = c$.Since $a, b > 1$,both $\log _a b$ and $\log _b a$ are positive real numbers.By the Arithmetic Mean-Geometric Mea

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$2$

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(C) Given $\log _a b + \log _b a = c$.Since $a, b > 1$,both $\log _a b$ and $\log _b a$ are positive real numbers.By the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality,for any two positive real numbers $x$ and $y$,$\frac{x+y}{2} \geq \sqrt{xy}$.Let $x = \log _a b$ and $y = \log _b a$.Then $\frac{\log _a b + \log _b a}{2} \geq \sqrt{\log _a b \cdot \log _b a}$.Since $\log _a b = \frac{1}{\log _b a}$,we have $\log _a b \cdot \log _b a = 1$.Thus,$\frac{c}{2} \geq \sqrt{1} = 1$.This implies $c \geq 2$.The smallest possible integer value of $c$ is $2$.

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

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