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1 solution
1) Any log with base x and number y can be written as [log(y)/log(x)]
2) if you write the numbers in question in the above manner you will see the numerator as [log2 +log3 + log4+....log100
3) denominator will be common log 100!
4) logA + logB + logC +...logZ = log( A B C .... Z)
5) therefore numerator = log (2 3 4 5..... 100) = log (100!)
6)therefore its log (100!)/ Log(100!) = 1