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Quantitative Aptitude

A.M Greater than or Equal to G.M

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Question

Find the minimum value of $l o g_{a} b + l o g_{b} a$ (where a and b are not equal to 1).

A

1

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B

2

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C

3

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D

4

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Solution

The correct option is B
2

$l o g_{a} b + l o g_{b} a = l o g_{a} b + \frac{1}{l o g_{a} b}$

Using $A . M . \geq G . M .$ -
$\frac{l o g_{a} b + \frac{1}{l o g_{a} b}}{2} \geq \sqrt{l o g_{a} b \times \frac{1}{l o g_{a} b}} \Rightarrow l o g_{a} b + \frac{1}{l o g_{a} b} \geq 2$

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