Determine the value of $$ \log_2 \log_2 \log_3 \log_3 27^3 $$.

Question

Accepted Answer

To solve $$ \log_2 \log_2 \log_3 \log_3 27^3 $$, we evaluate the expression step by step.

**Step 1: Simplify $$ \log_3 27^3 $$**  
Using the logarithmic property $$ \log_b (x^n) = n \cdot \log_b x $$:  
$$
\log_3 27^3 = 3 \cdot \log_3 27
$$  
Since $$ 27 = 3^3 $$, we know $$ \log_3 27 = 3 $$:  
$$
\log_3 27^3 = 3 \cdot 3 = 9
$$

**Step 2: Evaluate $$ \log_3 (\log_3 27^3) $$**  
Substitute $$ \log_3 27^3 = 9 $$:  
$$
\log_3 (\log_3 27^3) = \log_3(9)
$$  
Since $$ 9 = 3^2 $$, we know $$ \log_3(9) = 2 $$.

**Step 3: Evaluate $$ \log_2 (\log_2 (\log_3 (\log_3 27^3))) $$**  
Substitute $$ \log_3 (\log_3 27^3) = 2 $$:  
$$
\log_2 (\log_2 (2))
$$  
First, calculate $$ \log_2(2) = 1 $$:  
$$
\log_2 (\log_2 (2)) = \log_2(1)
$$  
Since $$ \log_b(1) = 0 $$ for any base $$ b > 0, b 
eq 1 $$:  
$$
\log_2 (\log_2 (2)) = 0
$$

Thus, the value of the expression is approximately **0**.

Answer

Answer

Answer