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Simplification of logarithmic alternating sum

CAT 2018 Slot 2 · Quantitative Ability · Medium · Logarithms

This is a medium Quantitative Ability question from the CAT 2018 Slot 2 paper. It tests Logarithms. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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The value of 1log⁡2100−1log⁡4100+1log⁡5100−1log⁡10100+1log⁡20100−1log⁡25100+1log⁡50100\frac{1}{\log_2 100} - \frac{1}{\log_4 100} + \frac{1}{\log_5 100} - \frac{1}{\log_{10} 100} + \frac{1}{\log_{20} 100} - \frac{1}{\log_{25} 100} + \frac{1}{\log_{50} 100} is

  1. A.

    0

  2. B.

    1 / 2

  3. C.

    -4

  4. D.

    10

Answer

B

Explanation

Using 1log⁡ab=log⁡ba\frac{1}{\log_a b} = \log_b a, the given expression becomes: log⁡1002−log⁡1004+log⁡1005−log⁡10010+log⁡10020−log⁡10025+log⁡10050\log_{100} 2 - \log_{100} 4 + \log_{100} 5 - \log_{100} 10 + \log_{100} 20 - \log_{100} 25 + \log_{100} 50

Combine using logarithmic properties: =log⁡100(2×5×20×504×10×25)= \log_{100} \left( \frac{2 \times 5 \times 20 \times 50}{4 \times 10 \times 25} \right)

Numerator =2×5×20×50=10000= 2 \times 5 \times 20 \times 50 = 10000. Denominator =4×10×25=1000= 4 \times 10 \times 25 = 1000.

NumeratorDenominator=100001000=10\frac{\text{Numerator}}{\text{Denominator}} = \frac{10000}{1000} = 10

So expression =log⁡10010=12= \log_{100} 10 = \frac{1}{2}.

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