\(\cfrac{1}{log_2-log_2-log_2\,16}\) = .........1/(log2 - log2 - log2

1.	\(\cfrac{1}{log_2-log_2-log_2\,16}\) = .........1/(log2 - log2 - log2 16)(A) 1 (B) 5 (C) 6 (D) 8
Answer» Correct option is (A) 1 \(\frac{1}{log_2\,log_2\,log_2\,16}\) \(=\frac{1}{log_2\,log_2\,log_2\,2^4}\) \(=\frac{1}{log_2\,log_2\,4\,log_2\,^2}\) \(=\frac{1}{log_2\,log_2\,2^2}\) \((\because log_2\,^2=1)\) \(=\frac{1}{log_2\,2\,log_2\,2}\) \(=\frac{1}{log_2\,2}\) \((\because log_2\,^2=1)\) = 1 Correct option is (A) 1

\(\cfrac{1}{log_2-log_2-log_2\,16}\) = .........1/(log2 - log2 - log2 16)(A) 1 (B) 5 (C) 6 (D) 8

Answer»

Correct option is (A) 1

\(\frac{1}{log_2\,log_2\,log_2\,16}\) \(=\frac{1}{log_2\,log_2\,log_2\,2^4}\)

\(=\frac{1}{log_2\,log_2\,4\,log_2\,^2}\)

\(=\frac{1}{log_2\,log_2\,2^2}\) \((\because log_2\,^2=1)\)

\(=\frac{1}{log_2\,2\,log_2\,2}\)

\(=\frac{1}{log_2\,2}\) \((\because log_2\,^2=1)\)

= 1

Correct option is (A) 1