Which one is smaller form the below:
$2$ or $({log}_{(e - 1)} 2 + {log}_{2} (e - 1))$

2

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({log}_{e - 1} 2 + {log}_{2} e - 1)

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Cannot be compared

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Both are equal

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Solution

The correct option is A $2$
$({log}_{e - 1} 2 + {log}_{2} e - 1)$
$= \frac{1}{{log}_{2} (e - 1)} + {log}_{2} (e - 1) \geq 2, [∵ x + \frac{1}{x} \geq 2 if x > 0]$
Hence $2$ is smaller.

Suggest Corrections

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