函数
已知2^a*3^b=2^c*3^d=6,求证(a-1)*(d-1)=(c-1)*(b-1)
2^a*3^b=2^c*3^d=6, 取常用对数得alg2+blg3=clg2+dlg3=lg2+lg3, ∴a=1+(1-b)lg3/lg2,d=1+(1-c)lg2/lg3, ∴(a-1)(d-1)=[(1-b)lg3/lg2][(1-c)lg2/lg3] =(b-1)(c-1).
答:(a-1)/(c-1)=(a-c+c-1)/(c-1) =(a-c)/(c-1)+1 =(d-b)lg3/((c-1)lg2)+1 (2^a*3^b=2^c*...详情>>
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