Question

Find the value of: $\frac{3+{\log }_{10}343}{2+\frac{1}{2}{\log }_{10}\left(\frac{49}{4}\right)+\frac{1}{2}{\log }_{10}\left(\frac{1}{25}\right)}$

Answer 1

$\frac{3+\log 343}{2+\frac{1}{2}\log \frac{49}{4}+\frac{1}{2}\log \left(\frac{1}{25}\right)}$

$=\frac{3+\log 7^3}{2+\frac{1}{2}\log 49-\frac{1}{2}\log 4-\frac{1}{2}\log 1-\frac{1}{2}\log 25}(\log \frac{a}{b}=\log a-\log b)$

$=\frac{3+3\log 7}{2+\frac{1}{2}\log 7^2-\frac{1}{2}{\log }^2-0-\frac{1}{2}{\log }^2}$

$=\frac{3(1+\log 7)}{2+\log 7-\log 2-\log 5}$

$\therefore \frac{3+\log 343}{2+\frac{1}{2}\log \frac{49}{4}+\frac{1}{2}\log \left(\frac{1}{25}\right)}=\frac{3(1+\log 7)}{2+\log 7-\log 2-\log 5}$.