若x1满足2x+2^x=5,x2满足2x+2log2(x-1)=5,则x1+x2=?2log2(x-1) 中间的2是底,我搞半天都不懂
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若x1满足2x+2^x=5,x2满足2x+2log2(x-1)=5,则x1+x2=?2log2(x-1)中间的2是底,我搞半天都不懂若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,则x1+x2=?2log2(x-1) 中间的2是底,我搞半天都不懂
若x1满足2x+2^x=5,x2满足2x+2log2(x-1)=5,则x1+x2=?
2log2(x-1) 中间的2是底,我搞半天都不懂
若x1满足2x+2^x=5,x2满足2x+2log2(x-1)=5,则x1+x2=?2log2(x-1) 中间的2是底,我搞半天都不懂
答：
x1满足：2x+2^x=5,2x1+2^(x1)=5
x2满足：2x+2log2(x-1)=5,2x2+2log2(x2-1)=5
设t=log2(x2-1)
则x2-1=2^t
所以：x2=1+2^t
所以：2(1+2^t)+2t=5
所以：2(t+1)+2^(t+1)=5
所以：
x1和t+1都是方程2x+2^x=5的解
所以：x1=t+1=log2(x2-1)+1=log2(2x2-2)
2x2-2=2^(x1)
2x2=2+2^(x1)
所以：
2x1+2x2=2x1+2+2^(x1)
=2x1+2+(5-2x1)
=7
解得：x1+x2=7/2
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