若x1满足2x+2的x方=5,x2满足2x+2log2(x-1)=5,求x1+x2

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若x1满足2x+2的x方=5,x2满足2x+2log2(x-1)=5,求x1+x2若x1满足2x+2的x方=5,x2满足2x+2log2(x-1)=5,求x1+x2若x1满足2x+2的x方=5,x2满

若x1满足2x+2的x方=5,x2满足2x+2log2(x-1)=5,求x1+x2
若x1满足2x+2的x方=5,x2满足2x+2log2(x-1)=5,求x1+x2

若x1满足2x+2的x方=5,x2满足2x+2log2(x-1)=5,求x1+x2
原来两方方程可化简得
2^(x－1)＝(5－2x)/2
log2(x－1)＝(5－2x)/2
设函数f1＝2^(x－1)
f2＝log2(x－1)
f3＝(5－2x)/2
则x1是f1与f3的交点
x2是f2与f3的交点
f1与f2是函数与反函数的关系,即关于y＝x对称
做草图,可得(x1＋x2)/2的值是f3与y＝x交点的横坐标
所以x1＋x2＝2.5

正确答案应该是7/2

令log2(x2-1)=t
∴x2=2^t+1
∴2(2^t+1)+2t=5
∴2^(t+1)+2(t+1)=5
∴t+1=x1
∴x1+x2
=t+1+2^t+1
=[2^(t+1)+2(t+1)]/2+1
=5/2+1
=7/2

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