对数计算题求解，谢谢

Question

如题所述

Answer 1

（1）
原式=(1/2)(lg32-lg49)-(4/3)x(1/2)lg8+(1/2)lg245
=(1/2)(lg2^5-lg7²)-(2/3)lg2³+(1/2)lg(5*49)
=(1/2)(5lg2-2lg7)-(2/3)x3lg2+(1/2)(lg5+lg49)
=(5/2)lg2-lg7-2lg2+(1/2)lg5+(1/2)lg7²
=(5/2-2)lg2+(1/2)lg5-lg7+(1/2)x2lg7
=(1/2)(lg2+lg5)
=(1/2)lg10
=1/2

(2)
原式=(lg5)²+(2/3)lg2³+lg5xlg20+(lg2)²
=(lg5)(lg5+lg20)+(2/3)x3lg2+(lg2)²
=(lg5)[lg(5x20)]+2lg2+(lg2)²
=(lg5)[lg100]+2lg2+(lg2)²
=(lg5)(lg10²)+2lg2+(lg2)²
=(lg5)(2lg10)+2lg2+(lg2)²
=(lg5)(2)+2lg2+(lg2)²
=2(lg5+lg2)+(lg2)²
=2lg10+(lg2)²
=2+(lg2)²
【此题可能印刷错误】