Question

log in baza 2018 din 1/2+log2018 din 2/3+log2018 din 3/4+ •••+ log 2018 din 2017/2018

Answer 1

$log_2_0_1_8\frac{1}{2}+log_2_0_1_8\frac{2}{3}+log_2_0_1_8\frac{3}{4}+...+log_2_0_1_8\frac{2017}{2018}=$

$log_2_0_1_8(\frac{1}{2}*\frac{2}{3}*\frac{3}{4}*...*\frac{2016}{2017}*\frac{2017}{2018})=$

$log_2_0_1_8\frac{1}{2018}=log_2_0_1_8(2018)^{-1}=-1*log_2_0_1_82018=-1*1=-1$