()若
满足2x+
=5,
满足2x+2
(x-1)=5,
+
=
(A)
(B)3 (C)
(D)4
()若
满足2x+
=5,
满足2x+2
(x-1)=5,
+
=
(A)
(B)3 (C)
(D)4
C
由题意
①
②
所以
,![]()
即2![]()
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