ABC26GN2003 · Logarithms

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

Evaluate : (i) $\log_{3} 27$ (ii) $\log_{7}\left(\frac{1}{343}\right)$ (iii) $\log_{100}(0.01)$ (iv) $\log_{8} 128$
Answer
Answer (as printed):
Explanation
(i) $\log_{3}27=\log_{3}3^{3}=3\log_{3}3=3$. $[\because \log_{3}3=1]$ (ii) $\log_{7}\left(\frac{1}{343}\right)=\log_{7}\left(\frac{1}{7^{3}}\right)=\log_{7}7^{-3}=-3\log_{7}7=-3$. (iii) Let $\log_{100}(0.01)=\log_{100}\left(\frac{1}{100}\right)=\log_{100}(100)^{-}=-1\log_{100}100=-1$. (iv) $\log_{8}128=\log_{2^{3}}\left(2^{7}\right)=\frac{7}{3}\log_{2}2=\frac{7}{3}$.

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