Question

已知函数f(x)=alog以2为底x的对数+blog3为底x的对数+2且f（1/2011）=4，则f(2011）的值为

Answer 1

0
f(x)=alog(2)x+blog(3)x+2 =algx/lg2 +blgx/lg3+2
=(a/lg2+b/lg3)lgx+2
f(1/2011)=(a/lg2+b/lg3)lg(1/2011)+2
=-(a/lg2+b/lg3)lg2011+2=4
所以(a/lg2+b/lg3)lg2011=-2
所以f(2011)==(a/lg2+b/lg3)lg2011+2=-2+2=0

Answer 2

f(1/2011)=alog2(1/2011)+blog3(1/2011)+2=-alog2(2011)-blog3(2011)+2=4
得alog2(2011)+blog3(2011)=-2
f(2011)=alog2(2011)+blog3(2011)+2=-2+2=0