$$\begin{gathered} 2) The value of \frac{\log 49\sqrt{7}+\log 25\sqrt{5}-\log 4\sqrt{2}}{\log 17.5} is \\ (a) 5 \\ (b) 2 \\ (c) \frac{5}{2} \\ (d) \frac{3}{2} \end{gathered}$$

Dear Student,

$$\begin{gathered} To Evaluate : \frac{\log 49\sqrt{7}+\log 25\sqrt{5}-\log 4\sqrt{2}}{\log 17.5} \\ solution : Using \log m+\log n=\log \left(mn\right) \\ \Rightarrow \frac{\log (7^2\sqrt{7}\times 5^2\sqrt{5})-\log 2^2\sqrt{2}}{\log 17.5} \\ Using \log m-\log n=\log \left(\frac{m}{n}\right) \\ \Rightarrow \frac{\log \left({\displaystyle \frac{7^2{7}^{{\displaystyle \frac{1}{2}}}\times 5^2{5}^{{\displaystyle \frac{1}{2}}}}{2^2{2}^{\frac{1}{2}}}}\right)}{\log 17.5} \\ =\frac{\log \left({\displaystyle \frac{7^{{\displaystyle \frac{5}{2}}}\times 5^{{\displaystyle \frac{5}{2}}}}{2^{\frac{5}{2}}}}\right)}{\log 17.5} \\ =\frac{\log ({\displaystyle \frac{7\times 5}{2}}{)}^{{\displaystyle \frac{5}{2}}}}{\log 17.5} \\ =\frac{{\displaystyle \frac{5}{2}} \cancel{\log (17.5)}}{\cancel{\log 17.5}} \\ =\frac{5}{2} \end{gathered}$$

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