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CAT 2018 Slot 2 — QA Question 23

LogarithmsEasy

1log2100-1log4100+1log5100-1log10100+1log20100-1log25100+1log50100?

Answer & solution

Correct answer: 1/2

  • A

    -4

  • B

    10

  • C

    0

  • 1/2

Solution

Easy

Flip each reciprocal log: 1log⁡a100=log⁡100a\dfrac{1}{\log_a 100}=\log_{100} a. The whole expression collapses into a single base-100100 log of a product/quotient of the numbers, which simplifies neatly.

1

Invert each term. Using 1log⁡ab=log⁡ba\dfrac{1}{\log_a b}=\log_b a.

1log⁡a100=log⁡100aso the sum=log⁡1002−log⁡1004+log⁡1005−log⁡10010 +log⁡10020−log⁡10025+log⁡10050\begin{aligned} &\frac{1}{\log_a 100}=\log_{100} a\\ &\text{so the sum}=\log_{100}2-\log_{100}4+\log_{100}5-\log_{100}10\\ &\qquad\qquad\ +\log_{100}20-\log_{100}25+\log_{100}50 \end{aligned}
2

Merge into one log. Plus signs multiply, minus signs divide.

=log⁡100 ⁣(2⋅5⋅20⋅504⋅10⋅25)=log⁡100 ⁣(100001000)=log⁡10010\begin{aligned} &=\log_{100}\!\left(\frac{2\cdot 5\cdot 20\cdot 50}{4\cdot 10\cdot 25}\right)\\ &=\log_{100}\!\left(\frac{10000}{1000}\right)=\log_{100}10 \end{aligned}
3

Evaluate. Since 100=102100=10^2, log⁡10010=12\log_{100}10=\tfrac12.

log⁡10010=12\begin{aligned} &\log_{100}10=\frac{1}{2} \end{aligned}
12\frac{1}{2}

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CAT 2018 Slot 2 QA Q23 — Logarithms solution | TheCATExam