ABC26GN2089 · Logarithms
Subject: General Aptitude · Chapter: Logarithms · Exam: 2002 · Marks: · Difficulty:
If $\log 2=0.3010$ and $\log 3=0.4771$, the value of $\log _{5} 512$ is
(a)2.870
(b)2.967
(c)3.876
(d)3.912
Answer
Explanation
$$\begin{aligned} \log _{5} 512 & =\frac{\log 512}{\log 5}=\frac{\log 2^{9}}{\log \left(\frac{10}{2}\right)}=\frac{9 \log 2}{\log 10-\log 2} \\ & =\frac{(9 \times 0.3010)}{1-0.3010}=\frac{2.709}{0.699}=\frac{2709}{699}=3.876 \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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