Question

已知函数f(x)=log2(4^x+1）+kx,(k是实数)是偶函数,求k的值.

Answer 1

f(x)=log2(4^x+1）+kx= log2(4^x+1）+ log2（2^kx）= log2[(2^2x+1）*2^kx]

由f(x)= f(-x)得log2[(2^2x+1）*2^kx]= log2[(2^2（-x）+1）*2^k（-x）]

由此得(2^2x+1）*2^kx=(2^2（-x）+1）*2^k（-x）

整理后可得k=-1.