Question

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

Answer 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^x<1- √3/2,
x>log2 (1+ √3/2)或 x<log2 (1- √3/2)

Answer 2

f(x)=log4(4^x+1)+kx是偶函数
∴log4(4^﹙﹣x﹚+1)-kx≡log4(4^x+1)+kx
∴log4[4^﹙﹣x﹚+1)／（(4^x+1)]-2kx≡0
∴log4[4^﹙﹣x)]-2kx≡0
∴﹣x-2kx≡0
∴2k=﹣1
∴k=﹣1／2
∴f(x)=log4(4^x+1)－x／2＞1
∴4^x+1＞4^﹙x／2+1﹚=4·4^﹙x／2﹚=4·2^x
令2^x=t，则t＞0
t²＋1＞4t
∴t＞2＋√3或0＜t＜2-√3
即2^x＞2＋√3或2^x＜2-√3
∴x＞log2﹙2＋√3﹚或x＜log2﹙2-√3﹚