Question

помогите, пожалуйста :с

Answer 1

$frac{3log_{27}288}{5log_32+2}=frac{3log_{left(3^3right)}288}{log_32^5+log_33^2}=\ =frac{frac33log_3288}{log_3(2^5cdot3^2)}=frac{log_3288}{log_3(32cdot9)}=frac{log_3288}{log_3288}=1.$

Answer 2

$frac{3log_{27}288}{5log_32+2} = frac{3log_{27}288}{log_32^5+log_33^2} = dfrac{ log_3288 }{log_3288} =1$