Evaluate, without any tables, $$ 3 \log 2+\log 20-\log 1 \cdot 6 $$

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To evaluate the expression:

$$
3 \log 2 + \log 20 - \log 1 \cdot 6
$$

let's break it down step by step.

1. **Simplify $$\log 1 \cdot 6$$:**
   
   $$
   \log 1 = 0 \quad \text{(since any logarithm of 1 is 0)}
   $$
   
   $$
   0 \cdot 6 = 0
   $$
   
   So, the expression simplifies to:
   
   $$
   3 \log 2 + \log 20
   $$

2. **Combine the logarithms using logarithm properties:**
   
   Recall that $$ a \log b = \log b^a $$ and $$ \log b + \log c = \log (b \cdot c) $$.

$$
   3 \log 2 = \log 2^3 = \log 8
   $$
   
   Now, add the two logarithms:
   
   $$
   \log 8 + \log 20 = \log (8 \cdot 20) = \log 160
   $$

Therefore, the simplified value of the original expression is:

$$
\log 160
$$

**Final Answer:**

Its value is the logarithm of 160, that is log 160.
The expression simplifies to $$ \log 160 $$.

Evaluate, without any tables,

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