log16(1/8)+log9(3√3)-log5(√5/5)
由换底公式
=lg(1/8)/lg16+lg(3√3)/lg9-lg(√5/5)/lg5
因为√3=3^(1/2),所以3√3=3*3^(1/2)=3^(1+1/2)=3^(3/2)
√5=5^(1/2),√5/5=5^(1/2-1)=5^(-1/2)
所以=lg2^(-3)/lg2^4+lg3^(3/2)/lg3^2-lg5^(-1/2)/lg5
=(-3)lg2/4lg2+(3/2)lg3/2lg3-(-1/2)lg5/lg5
=(-3)/4+(3/2)/2-(-1/2)/1
=-3/4+3/4+1/2
=1/2
2）
（12lg2+6lg3）/(6+3lg0.36+2lg8)
=（6lg4+6lg3)/(lg10^6+lg0.36^3+lg8^2)
=6lg12/lg(10^6*0.36^3*8^2)
=6lg12/lg(10^6*0.6^6*2^6)
=6lg12/6lg12
=1