ABC26GN2045 · Logarithms

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

$\log _{10} \frac{26}{51}+\log _{10} \frac{119}{91}-\log _{10} \frac{13}{32}-\log _{10} \frac{64}{39}$ is equal to
(a)0
(b)1
(c)2
(d)3
Answer
Answer (as printed): A
Explanation
$$\begin{aligned} & \log _{10} \frac{26}{51}+\log _{10} \frac{119}{91}-\log _{10} \frac{13}{32}-\log _{10} \frac{64}{39} \\ & =\left(\log _{10} \frac{26}{51}+\log _{10} \frac{119}{91}\right)-\left(\log _{10} \frac{13}{32}+\log _{10} \frac{64}{39}\right) \\ & =\log _{10}\left(\frac{26}{51} \times \frac{119}{91}\right)-\log _{10}\left(\frac{13}{32} \times \frac{64}{39}\right)=\log _{10} \frac{2}{3}-\log _{10} \frac{2}{3}=0 \end{aligned}$$

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

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