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

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.