Question

Помогите с первого по третий решить пж

Answer 1

1) 10^(3-2lg5)=(10^3)/10^(2lg5)=1000/10^(lg25)=1000/25=40;
2) log(2) 27-log(2) 9+log(2) (2/3)=log(2) (27:9*2/3)=log(2) 2=1;
3) 3^(1/(log(16) 81)=3^(log(81) 16)=3^(log(3^4) (2^4))=3^(log(3) 2)=2.