Question

请问下这题的一个式子怎么化简

函数f<X>=(log2[x]-1)/(log2[x]+1)若,f<4x1>+f<4x2>=1
怎么化简到log2[x1]+log2[x2]+logx[x1]log2[x2]=3
2是log的底数，[ ]是真数部分
要详解，谢谢了，我解了好久= =

Answer 1

应该是：log2[x1]+log2[x2]+log2[x1]log2[x2]=3 （你的第三个logx是打错了吧）

f<4x1>+f<4x2>=1→

(log2[4x1]-1)/(log2[4x1]+1)+(log2[4x2]-1)/(log2[4x2]+1)=1→把4和X分出来

(log2[4]+log2[x1]-1)/(log2[4]+log2[x1]+1)+(log2[4]+log2[x2]-1)/(log2[4]+log2[x2]+1)=1→计算各项
(log2[x1]+1)/(log2[x1]+3)+(log2[x2]+1)/(log2[x2]+3)=1→通分
[(log2[x1]+1)(log2[x2]+3)+(log2[x2]+1)(log2[x1]+3)]/[(log2[x1]+3)(log2[x2]+3)]=1→
分子分母相等并展开
3+log2[X2]+3log2[x1]+log2[x1]*log2[x2]+3+log2[x1]+3log2[x2]+log2[x2]*log2[x1]=9+log2[x1]*log2[x2]+3log2[x1]+3log2[x2]→左右相消就可以得到
log2[x1]+log2[x2]+log2[x1]log2[x2]=3
因为只能打字，图片我上传不了，你将就着看吧，明白思路就行。