0.01^(-1/2)-log0.5底8+3^(log3底2）+(lg2)^2+lg2*lg5+lg5=?
问题描述：
0.01^(-1/2)-log0.5底8+3^(log3底2）+(lg2)^2+lg2*lg5+lg5=?
答
0.01^(-1/2)-log0.5底8+3^(log3底2）+(lg2)^2+lg2*lg5+lg5=?
0.01 = 1/100
所以0.01 ^(1/2) = 1/10 = 0.1
所以 0.01^(-1/2) = 10
log0.5(8) = - log2(8) = -3
3^(log3底2 = 2
(lg2)^2+lg2*lg5+lg5=lg2*(lg2+lg5)+lg5=lg2*lg(2*5)+lg5=lg2+lg5=lg(2*5)=1
所以
原式 = 10 - (-3) + 2 + 1
= 10 + 3+ 2 + 1 = 16