ABC26GN2004 · Logarithms

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

Evaluate: (i) $\log _{7} 1=0$ (ii) $\log _{34} 34$ (iii) $36^{\log 64}$
Answer
Answer (as printed):
Explanation
(i) We know that $\log_{a}1=0$, so $\log_{7}1=0$. (ii) We know that $\log_{a}a=1$, so $\log_{34}34=1$. (iii) We know that $a^{\log_{a}x}=x$. Now, $36^{\log_{6}4}=6^{2(\log_{6}4)}=6^{\log_{6}(4^{2})}=6^{\log_{6}16}=16$.

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

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