问题
选择题
设函数f(x)=ax(a>0,a≠1),如果f(x1+x2+…+x2009)=8,那么f(2x1)×f(2x2)×…×f(2x2009)的值等于( )
A.32
B.64
C.16
D.8
答案
f(x1+x2+…+x2009)=8可得ax1+x2+…+x2009=8
f(2x1)×f(2x2)×…×f(2x2009)=a 2(x1+x2+…+x2009)=82=64
故选B.
设函数f(x)=ax(a>0,a≠1),如果f(x1+x2+…+x2009)=8,那么f(2x1)×f(2x2)×…×f(2x2009)的值等于( )
A.32
B.64
C.16
D.8
f(x1+x2+…+x2009)=8可得ax1+x2+…+x2009=8
f(2x1)×f(2x2)×…×f(2x2009)=a 2(x1+x2+…+x2009)=82=64
故选B.