分析:(1)利用對(duì)數(shù)的運(yùn)算法則把
lg25+lg8+lg5lg20+(lg2)2等價(jià)轉(zhuǎn)化為lg25+lg4+lg5(2lg2+lg5)+(lg2)
2,由經(jīng)能求出結(jié)果.
(2)由
log8a+log4b=,log8b+log4a2=7,知
log64a+log64b3=
,
log64b2+log64a6=7,故
ab3=64=85,a
3b=8
7,由此能求出a=64,b=8,從而得到ab的值.
解答:解:(1)
lg25+lg8+lg5lg20+(lg2)2=lg25+lg4+lg5(2lg2+lg5)+(lg2)
2=lg100+2lg5lg2+(lg5)
2+(lg2)
2=2+(lg2+lg5)
2=3;
(2)∵
log8a+log4b=,log8b+log4a2=7,
∴
log64a+log64b3=
,
log64b2+log64a6=7,
∴
ab3=64=85,
b
2a
6=64
7,即a
3b=8
7,
∴
=,即
=,a=8b,
∴a=64,b=8,
∴ab=64×8=512.
點(diǎn)評(píng):本題考查對(duì)數(shù)的性質(zhì)和運(yùn)算法則,是基礎(chǔ)題.解題時(shí)要認(rèn)真審題,仔細(xì)解答,注意合理地進(jìn)行等價(jià)轉(zhuǎn)化.