若lgA=a,lgB=b,lgC=c,则(lgA^2)*[lg(B/2]*(lg√C)=?

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若lgA=a,lgB=b,lgC=c,则(lgA^2)*[lg(B/2]*(lg√C)=?

(lgA^2)*[lg(B/2]*(lg√C)
=2logA*(lgB-lg2)*(1/2)lgC
=ac(b-lg2)