Question

log2(0,8)-log2(1,125)+log2(22,5)

Answer 1

$log_{2} 0.8-log_{2} 1.125+log_{2} 22.5=log_{2}( \frac{0.8}{1.125}*22.5)=log_{2}( \frac{4}{5} : \frac{9}{8} * \frac{45}{2} )=$$log_{2}( \frac{4}{5} * \frac{8}{9} * \frac{45}{2} )=log_{2}16=4$