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secondary 4 | A Maths
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Haii I really don't know how to solve this question on logarithms! Please help and thanks a lot! Any tips in trying to solve this is useful! :)
1/ log₂ 100! = log 2/ log 100!
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1/ log₁₀₀ 100! = log 100/ log 100!
so
sum
= 1/log 100! (log 2 + log 3 + ... + log 100)
= 1/log 100! [ log (2x3x4x ... x100) ]
= 1/log 100! (log 100!)
= 1
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