ABC26GN2041 · Logarithms

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

The value of $\log _{2} \log _{2} \log _{3} \log _{3} 27^{3}$ is
(a)0
(b)1
(c)2
(d)3
Answer
Answer (as printed): A
Explanation
$\log _{2} \log _{2} \log _{3}\left(\log _{3} 27^{3}\right)$ $=\log _{2} \log _{2} \log _{3}\left[\log _{3}\left(3^{3}\right)^{3}\right]=\log _{2} \log _{2} \log _{3}\left[\log _{3}(3)^{9}\right]$ $=\log _{2} \log _{2} \log _{3}\left(9 \log _{3} 3\right)=\log _{2} \log _{2} \log _{3} 9$ $\left[\therefore \log _{3} 3=1\right]$ $=\log _{2} \log _{2}\left[\log _{3}(3)^{2}\right]=\log _{2} \log _{2}\left(2 \log _{3} 3\right)$ $=\log _{2} \log _{2} 2=\log _{2} 1=0$.

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

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