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Convexity

For 0<a<x, ...

Question

For $0 < a < x$ , the minimum value of the function ${log}_{x} a + {log}_{a} x$ is:

1

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2

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- 2

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- \frac{1}{2}

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Solution

The correct option is D $2$
$y = l o g_{x} a + l o g_{a} x$
$= \frac{l n a}{l n x} + \frac{l n x}{l n a}$
$= \frac{l n a}{l n x} + \frac{1}{\frac{l n a}{l n x}}$
$= l o g_{x} a + \frac{1}{l o g_{x} a}$
Applying $A . M \geq G . M$
Hence
$\frac{l o g_{x} a + \frac{1}{l o g_{x} a}}{2} \geq \sqrt{l o g_{x} a . \frac{1}{l o g_{x} a}}$
$l o g_{x} a + \frac{1}{l o g_{x} a} \geq 2$
$y \geq 2$ .
Hence the minimum value of the function is 2.

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