Given that $\log 3 = 0.4771$ and $\log 2 = 0.3010$. We assume the logarithm is base 10. Problem (i): Calculate $\log 0.72$. Step 1: Express $0.72$ as a fraction and then in terms of prime factors. $$ 0.72 = \frac{72}{100} = \frac{2^3 \times 3^2}{10^2} $$ Step 2: Apply logarithm properties $\log \frac{a}{b} = \log a - \log b$ and $\log (xy) = \log x + \log y$. $$ \log 0.72 = \log \left(\frac{2^3 \times 3^2}{10^2}\right) = \log (2^3 \times 3^2) - \log (10^2) $$ $$ = \log (2^3) + \log (3^2) - \log (10^2) $$ Step 3: Apply the logarithm property $\log a^b = b \log a$. $$ = 3 \log 2 + 2 \log 3 - 2 \log 10 $$ Step 4: Substitute the given values and $\log 10 = 1$ (since we assume base 10). $$ = 3(0.3010) + 2(0.4771) - 2(1) $$ $$ = 0.9030 + 0.9542 - 2 $$ $$ = 1.8572 - 2 $$ $$ = -0.1428 $$ The value of $\log 0.72$ is $\boxed{-0.1428}$. --- Problem (ii): Calculate $\log \sqrt{56}$. Step 1: Rewrite the square root as an exponent