Question

Solve: $2{\log }_{10}{10}^3+3{\log }_{10}{10}^{-2}-\frac{1}{3}{\log }_{10}{5}^{-3}+\frac{1}{2}{\log }_{10}4$.

Answer 1

Now,

$2{\log }_{10}{10}^3+3{\log }_{10}{10}^{-2}-\frac{1}{3}{\log }_{10}{5}^{-3}+\frac{1}{2}{\log }_{10}4$

$=(2\times 3){\log }_{10}10+(3\times -2){\log }_{10}10-(\frac{1}{3}\times -3){\log }_{10}5+{\log }_{10}\sqrt{4}$ [Using rule of logarithm]

$=6-6+{\log }_{10}5+{\log }_{10}2$

$={\log }_{10}10$

$=1$.