已知函数f(x)=log4(4^x +1) g(x)=(k-1)x 记F(x)=f(x) - g(x),且F(x)为偶函数,求常数K

F(x)为偶函数,F(x)=F(-x)
log4(4^x +1)-(k-1)x =log4(4^(-x) +1)+(k-1)x
log4[(4^x +1)/(4^(-x) +1)]=2(k-1)x
(4^x +1)/(4^(-x) +1)=(4^x +1)/[(1+4^x)/4^x]=4^x
原式=log4 4^x=2(k-1)x,即x=2(k-1)x,2(k-1)=1 k=3/2