已知2logbx=logax+logcx(x≠1)求证C^2=bc^logab

已知2logbx=logax+logcx(x≠1)求证C^2=bc^logab

原式可以变为
2lgx/lgb=lgx/lga+lgx/lgc
2/lgb=1/lga+1/lgc=(lga+lgc)/(lgalgc)
2/lgb=lg(ac)/(lgalgc)
2lgalgc=lgblgac
lgalgc²=lgblgac
lgc²=(lgblgac)/lga
c²=10^[(lgblgac)/lga]
=10^[lgb(lga+lgc)/lga]
=10^[lgb(1+loga c)]
=10^(lgb+lgbloga c)
=10^lgb*10^(lgbloga c))
=b*10^(lgblgc/lga))
=b*10^(lgc*logab)
=b*(10^lgc)^(log ab)
=b*c^(logab)

2lgx/lgb=lgx/lga+lgx/lgc
2/lgb=1/lga+1/lgc
两边乘lgblgc
2lgc=lgb(lgc/lga+1)=lgb(lgc+lga)/lga
2lgc=lgb/lga*lg(ac)
lgc²=lg(ac)^loga(b)
c²=(ac)^loga(b)
求证的是不是有误？
追问：不是，我在百度查的都是这个求证，所以才提问嘛