已知函数f(x)=log4(4^x+1)+kx是偶函数,解不等式f（x)＞1

偶函数f(-x)=f(x),
log4[4^(-x)+1]-kx=log4(4^x+1)+kx,
log4 { [4^(-x)+1]/(4^x+1) }=2kx,
log4 1/4^x =2kx,
-x=2kx,
k=-1/2,
f（x)＞1,--> log4(4^x+1)-x/2 >1,
log4(4^x+1) >x/2 +1,
4^x+1>4^(x/2 +1),
(2^x)²-4*2^x +1>0,
2^x>1+ √3/2或2^xlog2 (1+ √3/2)或 x