Huge log.....

Evaluate

i = 1 99999 log 10 i i + 1 \left| \sum _{ i=1 }^{ 99999 }{ \log _{ 10 }{ \frac { i }{ i+1 } } } \right|

where |x| is the absolute value of x.


The answer is 5.

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1 solution

Christian Daang
Nov 4, 2014

If i = {1,2,3,4,5, ... , 99999} ,

The expression will be:

|log [base 10] (1/2) + log [base 10] (2/3) + ... + log [base 10] (99999/100000)|

= |log [base 10] [ (1/2) * (2/3) * ... * (99999/100000)]|

= |log [base 10] [1/100000]|

= |log [base 10] [(10)^(-5)]|

= |(-5) log [base 10] [10]|

= |(-5)(1)|

= |-5|

= 5

Final answer: 5

Alternately in step 2 you could expand each logarithm in the form of (log a - log b) to cancel out all the terms except the first and the last, the first being log 1. The answer would remain the same though.

Arko Bhaumik - 6 years, 6 months ago

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