若x1满足2x+2^x=5,x2满足2x+2log2(x-1)=5
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若x1满足2x+2^x=5,x2满足2x+2log2(x-1)=5若x1满足2x+2^x=5,x2满足2x+2log2(x-1)=5若x1满足2x+2^x=5,x2满足2x+2log2(x-1)=5令
若x1满足2x+2^x=5,x2满足2x+2log2(x-1)=5
若x1满足2x+2^x=5,x2满足2x+2log2(x-1)=5
若x1满足2x+2^x=5,x2满足2x+2log2(x-1)=5
令t=log2(x-1),则x=2^t+1
故2x+2log2(x-1)=5等价于2(2^t+1)+2t=5,
即2^(t+1)+2(t+1)=5
因为函数f(x)=2x+2^x是增函数,所以f(x)=5的根仅有一个
即t+1=x1,即
log2(x2-1)+1=x1,而2x2+2log2(x2-1)=5
x2=5/2-log2(x2-1)
故x1+x2=log2(x2-1)+1+x2=5/2+1=7/2
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