分析:(1)利用對(duì)數(shù)的基本性質(zhì)將第一項(xiàng)中各對(duì)數(shù)化為同底的進(jìn)行轉(zhuǎn)化變形,進(jìn)而求出其值,再利用對(duì)數(shù)的求解求出該式的值;
(2)利用指數(shù)冪的運(yùn)算性質(zhì)進(jìn)行求解化簡(jiǎn)是解決本題的關(guān)鍵.注意應(yīng)用同底數(shù)冪乘法的運(yùn)算性質(zhì).
解答:解:(1)原式=
(log223+log233)(log32+log322)-log2=
(log23+log2 3)(log32+log3 2)+=
××log23×log32+=
+=2.
(2)所化簡(jiǎn)的式子=
[a-ba-b-2×(-) a-1×(-) ]2=
(a-1+b1+1)2=
a-b4.,
代入a=8,b=-2,
計(jì)算得出原式的值為
(23)-×(-2)4=×16=4.
點(diǎn)評(píng):本題考查對(duì)數(shù)的運(yùn)算性質(zhì),考查對(duì)數(shù)式的運(yùn)算,關(guān)鍵要將不同底數(shù)的化為同底數(shù)的式子進(jìn)行轉(zhuǎn)換求值,指數(shù)式的運(yùn)算要用到同底數(shù)冪的運(yùn)算性質(zhì),注意先化簡(jiǎn)再求值.