ABC26GN2053 · Logarithms
Subject: General Aptitude · Chapter: Logarithms · Exam: 2010 · Marks: · Difficulty:
If $\log _{10} 2=a$ and $\log _{10} 3=b$, then $\log _{5} 12$ equals
(a)$\frac{a+b}{1+a}$
(b)$\frac{2 a+b}{1+a}$
(c)$\frac{a+2 b}{1+a}$
(d)$\frac{2 a+b}{1-a}$
Answer
Explanation
$\log _{5} 12=\log _{5}(3 \times 4)=\log _{5} 3+\log _{5} 4=\log _{5} 3+2 \log _{5} 2$ $$\begin{aligned} & =\frac{\log _{10} 3}{\log _{10} 5}+\frac{2 \log _{10} 2}{\log _{10} 5}=\frac{\log _{10} 3}{\log _{10} 10-\log _{10} 2}+\frac{2 \log _{10} 2}{\log _{10} 10-\log _{10} 2} \\ & =\frac{b}{1-a}+\frac{2 a}{1-a}=\frac{2 a+b}{1-a} . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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