Question

1)log(81) 9= 2)log(2) log(4) 256= 3)log(2) 10 - log(2) 5 + log(5) 8= 4)log(3) 7 log(4) 81 log(7) 2=

Answer 1

$log_{81}{9}=log_{9^2}{9}= \frac{1}{2}log_9{9}=1/2; log_2 log_4{256}=log_2log_4{4^4}=log_2{4}=$$=log_2{2^2}=2; log_2{10}-log_2{5}+log_5{8}=log_2{ \frac{10}{5} +log_5{8}=log_2{2}+log_5{8}=$$=1+log_5{8}; log_3{7}*log_4{81}*log_7{2}= \frac{log_2{7}}{log_2{3}}* \frac{4}{2}log_2{3}* \frac{log_2{2}}{log_2{7}}= \frac{4}{2}=2$