@ yxue :
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@yxue:@yxue:@yxue:证明:2/logaab+2/logbab=2(1)只需证明:1/logaab+1/logbab=1(2)为此,令:A=logaab,a^A=ab取常用对数lg:Al
@ yxue :
@ yxue :
@ yxue :
证明:
2/loga ab +2/logb ab = 2 (1)
只需证明:
1/loga ab + 1/logb ab = 1 (2)
为此,令:
A=loga ab ,a^A=ab 取常用对数lg:Alga=lgab,A=lgab/lga (实际上用了换底公式)
B=logb ab ,b^B=ab 取常用对数lg:Blgb=lgab,B=lgab/lgb
(2)式变成:1/A + 1/B = lga/lgab + lgb/lgab=(lga+lgb)/lgab=1
即:1/loga ab + 1/logb = 1 (2)
成立,也即(1)式成立.证毕.
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