ABC26GN2047 · Logarithms

Subject: General Aptitude · Chapter: Logarithms · Exam: · Marks: · Difficulty:

The value of $\log _{10} 1 \frac{1}{2}+\log _{10} 1 \frac{1}{3}+\cdots$ up to 198 terms is equal to
(a)0
(b)2
(c)10
(d)100
Answer
Answer (as printed): B
Explanation
$\log _{10} 1 \frac{1}{2}+\log _{10} 1 \frac{1}{3}+\cdots$ upto 198 $$\begin{aligned} & \text { terms }=\log _{10}\left(1 \frac{1}{2} \times 1 \frac{1}{3} \times \cdots \times 1 \frac{1}{199}\right) \\ & \begin{aligned} =\log _{10}\left(\frac{3}{2} \times \frac{4}{3} \times \cdots \times \frac{200}{199}\right) & =\log _{10}\left(\frac{200}{2}\right) \\ & =\log _{10} 100=\log _{10} 10^{2}=2 \log _{10} 10=2 \end{aligned} \end{aligned}$$

Explanation as extracted from the printed page; notation may be imperfect.

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