8^x-4^(x+1/2)-4^(x+m/2)+2^(x+1

8^x - 4^(x+1/2) - 4^(x+m/2) + 2^(x+1+m) + 2^(x+3) - 16 = 0
==> (2^x)^3 - 2(2^x)^2 - (2^m)(2^x)^2 + 2(2^m)(2^x) + 8(2^x) - 16 = 0
令 y = 2^x, n = 2^m, 則
y^3 - 2y^2 - ny^2 + 2ny + 8y - 16 = 0
==> (y - 2)y^2 - ny(y - 2) + 8(y - 2) = 0
==> (y - 2)(y^2 - ny + 8) = 0
如果所有根之中恰有兩根相等，則
y = 2 或 (y^2 - ny + 8) = 0 的判別式是 0
==> 2^2 - 2n + 8 = 0 或 n^2 = 4*8 = 2^5
==> 2^m = 6 或 2^(2m) = 2^5
==> m = (log 6)/(log 2) (拒納, 不是有理數) 或 m = 5/2

答案：m = 5/2

螞蟻兄：
這裡有點筆誤

4=2+2^m－4 ==> 2^m=4+4-2=6 , m= log(2,6) 不是有理數

設m為有理數，指數方程式8^x－4^(x+1/2)－4^(x+m/2)+2^(x+1+m)+2^(x+3)
－16=0的所有根之中恰有兩根相等，則m=???????
Sol
設 y=2^x>0
8^x－4^(x+1/2)－4^(x+m/2)+2^(x+1+m)+2^(x+3)－16=0
y^3－2y^2－(2^m)y^2+2^(1+m)y+8y－16=0
y^3－(2+2^m)y^2+(2^(1+m)+8)y－16=0
再設三根為p，p，q
p+p+q=2+2^m，p^2+pq+pq=2^(1+m)+8，p^2q=16
q=2+2^m－2p
p^2+2(2+2^m－2p)=2^(1+m)+8
p^2+4+2^(1+m)－4p=2^(1+m)+8
p^2－4p－4=0
(p－2)^2=0
p=2
q=4
4=2+2^m－4
m=1