若x1满足2x+2x=5,x2满足2x+2log2(x-1)=5,x1+x2=(  ) A.52 B.3 C.72 D.4

问题描述:

若x1满足2x+2x=5,x2满足2x+2log2(x-1)=5,x1+x2=(  )
A.

5
2

B. 3
C.
7
2

D. 4

由题意2x1+2x1=5
2x2+2log2(x2-1)=5  ②
所以2x1=5−2x1
x1=log2(5-2x1)   即2x1=2log2(5-2x1
令2x1=7-2t,代入上式得7-2t=2log2(2t-2)=2+2log2(t-1)
∴5-2t=2log2(t-1)与②式比较得t=x2
于是2x1=7-2x2
即x1+x2=

7
2

故选C