Question

高一数学对数计算

Answer 1

原式=lg2(lg2+lg50)+lg5²
=lg2lg(2×50)+2lg5
=lg2lg100+2lg5
=2lg2+2lg5
=2(lg2+lg5)
=2

Answer 2

(lg2)^2+lg2*lg50+lg25
=(lg2)^2+lg2(lg5*10)+lg25
=(lg2)^2+lg2(lg5+lg10)+lg25此处lg10=1
=(lg2)^2+lg2*lg5+lg2+lg25
=lg2(lg2+lg5)+lg2+lg25
=lg2+lg2+lg25
=lg(2*2*25)
=lg100
=2