Given log base 3 = 0.4771 and log 2 = 0.3010, evaluate log 162 + log 36 - log 54

The given expression is $\log 162 + \log 36 - \log 54$.
This can be written as:

$$\log \left(\frac{162 \times 36}{54}\right)$$

First, simplify the fraction $\frac{36}{54}$. Both numbers are divisible by 18:

$$\frac{36}{54} = \frac{36 \div 18}{54 \div 18} = \frac{2}{3}$$

Now substitute this back into the expression:

$$\log \left(162 \times \frac{2}{3}\right)$$

Perform the multiplication:

$$162 \times \frac{2}{3} = \frac{162 \times 2}{3} = \frac{324}{3} = 108$$

So the expression simplifies to $\log 108$.

$$108 = 2 \times 54$$ $$54 = 2 \times 27$$ $$27 = 3 \times 9$$ $$9 = 3 \times 3$$

Therefore, $108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$.

Using the property $\log(ab) = \log a + \log b$:

$$\log 108 = \log (2^2 \times 3^3) = \log(2^2) + \log(3^3)$$

Using the property $\log(a^b) = b \log a$:

$$\log(2^2) + \log(3^3) = 2 \log 2 + 3 \log 3$$

Given $\log 3 = 0.4771$ and $\log 2 = 0.3010$.

$$2 \log 2 + 3 \log 3 = 2(0.3010) + 3(0.4771)$$ $$= 0.6020 + 1.4313$$ $$= 2.0333$$

The final answer is 2.0333.