若x1满足2x+2^x=5,x2满足2x+2log2(x+1)=5,求x1+x2=多少?2X+2^x=5x+2^(x-1)=2.52X+2log2(x-1)(2是底数）=5x+log2(x-1)=2.5其中y=2^(x-1)与y=log2(x-1)关于y=x-1对称y=2.5-x与前面指数对数曲线交点即为x1,x2.x1+x2为y=2.5-x,y=x-1交点

Question

若x1满足2x+2^x=5,x2满足2x+2log2(x+1)=5,求x1+x2=多少?
2X+2^x=5
x+2^(x-1)=2.5
2X+2log2(x-1)(2是底数）=5
x+log2(x-1)=2.5
其中y=2^(x-1)与y=log2(x-1)关于y=x-1对称
y=2.5-x与前面指数
对数曲线交点即为x1,x2.
x1+x2为y=2.5-x,y=x-1交点横坐标2倍即3.5
对数曲线交点即为x1,x2.
x1+x2为y=2.5-x,y=x-1交点横坐标2倍即3.5

Answer 1

第一个式子x+2^(x-1)=2.5,即2^(x-1)=2.5-x若令y=2^(x-1),则y=2.5-x,x1就是y=2^(x-1),y=2.5-x交点横坐标同理x+log2(x-1)=2.5,log2(x-1)=2.5-x,若令y=log2(x-1),则y=2.5-x,x2就是y=log2(x-1),y=2.5-x交点横坐标y=2^(x-...