logaXlogbXlogcX成等差数列求证:c^2=(ac)^logaBlogaXlogbXlogcX成等差数列求证:c^2=(ac)^logaB

Question

log aX log bX log cX 成等差数列 求证:c^2=(ac)^log aB
log aX log bX log cX 成等差数列 求证:c^2=(ac)^log aB

Answer 1

由题意,log aX log bX log cX 成等差数列,则
2logbX=logaX+logcX,即2/logxB=1/logxA+1/logxC
通分,十字相乘得2logxA*logxC=logxB*logxAC
2=(logxB*logxAC)/(logxA*logxC),利用换底公式有
2=(lgB*lgAC)/(lgA*lgC)=logaB*logcAC=logaB*(logcA+1)
由结论c^2=(ac)^log aB,c^2=a^logaB*c^logaB=b*c^logaB
所以c^(2-logaB)=b
2-logaB=logaB*(logcA+1)-logaB=logaB*logcA=lgB/lgA*lga/lgc=lgB/lgC=logcB
因为c^logcB=b,所以c^(2-logaB)=b成立,则结论成立.