若log以a为底c＋log以b为底c＝0（c≠0）,则ab＋c－abc

做这一题,你首先要知道一个换底公式log(a)b=log(n)b/log(n)a(带括号的是底数,不带括号的是真数!)
log(a)c = log(c)c / log(c)a = 1/log(c)a
log(b)c = log(c)c / log(c)b = 1/log(c)b
所以原式 1/log(c)a + 1/log(c)b = 0
通分后：(log(c)b + log(c)a) / (log(c)a * log(c)b) = 0
所以 log(c)b + log(c)a = 0 所以log(c)ab = 0 所以 ab = 1;
所以 ab + c - abc = 1+c-c = 1;
写的很细!
有什么问题继续讨论!

∵log(a)c+log(b)c=0
=>1/[log(c)a]+1/[log(c)b]=0
=>-log(c)b=log(c)a
=>log(c)a+log(c)b=0
=>log(c)(ab)=0
=>ab=1
∴ab＋c－abc=1