已知函数f(x)=log4(4^x+1)+kx是偶函数,解不等式f（x)＞1
<p>问题：已知函数f(x)=log4(4^x+1)+kx是偶函数,解不等式f（x)＞1<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">梁宏伟的回答：<div class="content-b">网友采纳 偶函数f(-x)=f(x), log4-kx=log4(4^x+1)+kx, log4{/(4^x+1)}=2kx, log41/4^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
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