loga c*logc a

a、b、c都为正数,且都不为1,求log_a(b)+2log_b(c)+4log_c(a)的最小值.log_x(y)=lg(y)/lg(x)（换底公式,lg为常用对数）设p=lga,q=lgb,r=l

1、logaC×logcA=logaC×(1/logaC)=12、log23×log34×log45×log52=[1/(log32)]×2log32×[1/(log54)]×log52=2×[1/2

解析:∵logac·logbc=4,∴1=4logca·logcb≤4·=logcc^2=logc2ab,即logc2ab≥1.又a,b,c均大于1,∴logcab＞0∴cab≥1.∴ab≥c.☆⌒_

没了不可以直接变成a/b两者完全不同

证明：logaC*logbC=4,即1/4=logca*logcb≤[（logca+logcb）/2]^2,即1≤(logcab)^2又∵abc均大于1∴logcab≥1∴ab≥c

log(a)b.log(b)c.log(c)a=1证明：∵log(a)b.log(b)c.log(c)a=log(a)b.(log(a)c/log(a)b).(1/log(a)c)=1∴log(a)b

a^logcB=b^logcAln(a^logcB)=ln(b^logcA)logcB*lna=logcAlnb(lnb*lna)/lnc=(lnb*lna)/lnc,显然成立再问：a^logcB=b