The value of $\log_2 \log_2 \log_3 \log_3 27^3$ is

Concept & Formula This problem tests the sequential evaluation of nested logarithms, working from the rightmost (innermost) term to the left. The essential properties used are: $$\log_a (b^n) = n \log_a b$$ $$\log_a a = 1$$ $$a^0 = 1 \implies \log_a 1 = 0$$ Step-by-Step Solution * **Given:** $\log_2 \log_2 \log_3 \log_3 (27^3)$ * **Simplify the Innermost Argument:** First, express 27 as a power of 3. $$27^3 = (3^3)^3$$ * Apply the power of a power rule $(a^m)^n = a^{mn}$. $$(3^3)^3 = 3^9$$ * **Evaluate the First Logarithm (Rightmost):** $$\log_3 (3^9) = 9$$ * **Substitute and Evaluate the Second Logarithm:** Replace the evaluated part in the main expression. $$\log_2 \log_2 \log_3 (9)$$ * Since $9 = 3^2$, evaluate $\log_3 9$. $$\log_3 (3^2) = 2$$ * **Substitute and Evaluate the Third Logarithm:** $$\log_2 \log_2 (2)$$ * Since $2^1 = 2$, evaluate $\log_2 2$. $$\log_2 2 = 1$$ * **Evaluate the Final Logarithm (Leftmost):** $$\log_2 (1)$$ * Since any non-zero base raised to the power of 0 is 1, $\log_a 1 = 0$. $$0$$ Exam Strategy & Shortcut Solve strictly right-to-left. Recognize that $27^3$ is $3^9$. Mentally step through the chain: 1. $\log_3$ of $3^9$ is 9. 2. $\log_3$ of 9 is 2. 3. $\log_2$ of 2 is 1. 4. $\log_2$ of 1 is 0. This structured mental chain bypasses the need for any complex written algebra. Common Pitfall A common trap is incorrectly calculating the initial $27^3$ term by doing something like $3 \times 3 = 9$ or confusing it with $27 \times 3 = 81$. Ensure you use the correct exponent rule: $(3^3)^3 = 3^9$. Another mistake is stopping at the value of 1, forgetting the outermost $\log_2$ which turns the 1 into a 0. Final Answer **Therefore, the correct answer is 0.**