$\begin{align}  & \text{Show without using tables that } \\ & (i)\text{ }2{{\log }_{10}}6+3{{\log }_{10}}2={{\log }_{10}}288 \\ & \text{Solution} \\ & 2{{\log }_{10}}6+3{{\log }_{10}}2={{\log }_{10}}{{6}^{2}}+{{\log }_{10}}{{2}^{3}} \\ & ={{\log }_{10}}36+{{\log }_{10}}8 \\ & ={{\log }_{10}}(36\times 8) \\ & ={{\log }_{10}}(288) \\ & (ii)\text{ }{{\log }_{10}}19\approx \tfrac{1}{2}(2{{\log }_{10}}6+1) \\ & \text{Hence, show that }{{\log }_{10}}17\approx \tfrac{1}{2}(2{{\log }_{10}}6+3{{\log }_{10}}2) \\ & \text{Solution} \\ & \text{lo}{{\text{g}}_{10}}{{19}^{2}}={{\log }_{10}}(361) \\ & 2{{\log }_{10}}19\approx {{\log }_{10}}(360) \\ & 2{{\log }_{10}}19\approx {{\log }_{10}}(36\times 10) \\ & 2{{\log }_{10}}19\approx {{\log }_{10}}36+{{\log }_{10}}10 \\ & 2{{\log }_{10}}19\approx {{\log }_{10}}{{6}^{2}}+1 \\ & {{\log }_{10}}19\approx \tfrac{1}{2}(2{{\log }_{10}}6+1) \\ &  \\ & {{\log }_{10}}{{17}^{2}}={{\log }_{10}}(289) \\ & 2{{\log }_{10}}17={{\log }_{10}}289 \\ & 2{{\log }_{10}}17\approx \log 288 \\ & 2{{\log }_{10}}17\approx 2{{\log }_{10}}6+3{{\log }_{10}}2 \\ & {{\log }_{10}}17\approx \tfrac{1}{2}(2{{\log }_{10}}6+3{{\log }_{10}}2) \\ & {{\log }_{10}}19-{{\log }_{10}}17\approx \tfrac{1}{2}(2{{\log }_{10}}6+1)-\tfrac{1}{2}(2{{\log }_{10}}6+3{{\log }_{10}}2) \\ & {{\log }_{10}}\frac{19}{17}\approx \frac{1}{2}\left( 2{{\log }_{10}}6+1-2{{\log }_{10}}-3{{\log }_{10}}2 \right) \\ & {{\log }_{10}}\frac{19}{17}\approx \frac{1}{2}\left[ 1-3{{\log }_{10}}2 \right] \\ & {{\log }_{10}}\frac{19}{17}\approx \frac{1}{2}\left[ {{\log }_{10}}10-{{\log }_{10}}{{2}^{3}} \right] \\ & {{\log }_{10}}\frac{19}{17}\approx \frac{1}{2}{{\log }_{10}}\frac{10}{8} \\ & {{\log }_{10}}\frac{19}{17}\approx \frac{1}{2}{{\log }_{10}}\frac{5}{4} \\ & {{\log }_{10}}\frac{19}{17}\approx {{\log }_{10}}{{\left( \frac{5}{4} \right)}^{\tfrac{1}{2}}} \\ & {{\log }_{10}}\frac{19}{17}\approx {{\log }_{10}}\frac{\sqrt{5}}{2} \\ & \frac{19}{17}\approx \frac{\sqrt{5}}{2} \\ & \sqrt{5}\approx \frac{38}{17} \\\end{align}$