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Question

The least integer greater than log215log162log316 is equal to

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Solution

let y=log215log162log316
y=log215(1)log62(1)log36[log1/ab=logab & loga1b=logab]
=log15log2×log2log6×log6log3[logba=logalogb]
=log315[logba=logalogb]=log3(3×5)=log33+log35[log(ab)=loga+logb]
=1+log35[logaa=1]
Since, log5>log3
Therefore, y=1+log35>2
and the smallest possible integer is 3
Ans: 3

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