Question

Calculate the value of ${\log }_{2}32+{\log }_{4}49+{\log }_{8}125-{\log }_{2}1120$

Answer 1

$[Use{\log }_{a^n}b={\log }_a{b}^{1/n}$

${\log }_{2}32+{\log }_{4}49+{\log }_{8}125-{\log }_{2}1120$

$={\log }_2\left(32\right)+{\log }_{2^2}\left(49\right)+{\log }_{2^3}\left(125\right)-{\log }_{2}1120$

$={\log }_2\left(32\right)+{\log }_2(49{)}^{1/2}+{\log }_2(125{)}^{1/3}-{\log }_{2}1120$

$={\log }_2\frac{32\times 7\times 5}{1120}={\log }_{2}1=0$