3. Use logarithms only to evaluate, correct to 4 decimal places: $$ \left(\frac{12.78 \sin 33.45^\circ}{(9.063)^2}\right)^{1/3} $$ Let $X = \left(\frac{12.78 \sin 33.45^\circ}{(9.063)^2}\right)^{1/3}$. Taking logarithms to base 10: $$ \log X = \frac{1}{3} \left( \log(12.78) + \log(\sin 33.45^\circ) - 2 \log(9.063) \right) $$ Step 1: Find the logarithm of each term. Using logarithm tables: $\log(12.78) = 1.1065$ $\sin 33.45^\circ = 0.5512$ $\log(\sin 33.45^\circ) = \log(0.5512) = \bar{1}.7413$ $\log(9.063) = 0.9572$ Step 2: Substitute the logarithm values into the equation for $\log X$. $$ \log X = \frac{1}{3} (1.1065 + \bar{1}.7413 - 2 \times 0.9572) $$ $$ \log X = \frac{1}{3} (1.1065 + (-1 + 0.7413) - 1.9144) $$ $$ \log X = \frac{1}{3} (1.1065 - 0.2587 - 1.9144) $$ $$ \log X = \frac{1}{3} (0.8478 - 1.9144) $$ $$ \log X = \frac{1}{3} (-1.0666) $$ Step 3: Divide by 3. $$ \log X = -0.355533... $$ Step 4: Convert the negative logarithm to characteristic-mantissa form. $$ -0.3555 = -1 + 0.6445 $$ So, $\log X = \bar{1}.6