Question

Solve: ${\log }_{10}\left(\frac{12}{5}\right)+{\log }_{10}\left(\frac{25}{21}\right)-{\log }_{10}\left(\frac{2}{7}\right)$

Answer 1

We have,

${\log }_{10}\left(\frac{12}{5}\right)+{\log }_{10}\left(\frac{25}{21}\right)-{\log }_{10}\left(\frac{2}{7}\right)$

$\Rightarrow {\log }_{10}12-{\log }_{10}5+{\log }_{10}25-{\log }_{10}21-{\log }_{10}2+{\log }_{10}7$

$\Rightarrow {\log }_{10}(3\times 4)-{\log }_{10}5+{\log }_{10}(5\times 5)-{\log }_{10}(3\times 7)-{\log }_{10}2+{\log }_{10}7$

$\Rightarrow {\log }_{10}3+{\log }_{10}4-{\log }_{10}5+{\log }_{10}5+{\log }_{10}5-{\log }_{10}3-{\log }_{10}7-{\log }_{10}2+{\log }_{10}7$

$\Rightarrow {\log }_{10}4+{\log }_{10}5-{\log }_{10}2$

$\Rightarrow {\log }_{10}(2\times 2)+{\log }_{10}5-{\log }_{10}2$

$\Rightarrow {\log }_{10}2+{\log }_{10}2+{\log }_{10}5-{\log }_{10}2$

$\Rightarrow {\log }_{10}2+{\log }_{10}5$

$\Rightarrow {\log }_{10}10$

$\Rightarrow 1$

Hence, this is the answer.