a,b,c為異於1的正時數，若(1/loga)+(1/logb)+(1/logc)=1/(loga+logb+logc)
求(ab-1)(bc-1)(ca-1)之值
對數
版主: thepiano
Re: 對數
令 loga = x，logb = y，logc = z
1/x + 1/y + 1/z = 1/(x + y + z)
(yz + zx + xy)(x + y + z) - xyz = 0
x^2y + y^2z + z^2x + xy^2 + yz^2 + zx^2 + 2xyz = 0
(x + y)(y + z)(z + x) = 0
x + y = log(ab)
ab = 10^(x + y)
故 ab = 1 or bc = 1 or ca = 1
(ab - 1)(bc - 1)(ca - 1) = 0
1/x + 1/y + 1/z = 1/(x + y + z)
(yz + zx + xy)(x + y + z) - xyz = 0
x^2y + y^2z + z^2x + xy^2 + yz^2 + zx^2 + 2xyz = 0
(x + y)(y + z)(z + x) = 0
x + y = log(ab)
ab = 10^(x + y)
故 ab = 1 or bc = 1 or ca = 1
(ab - 1)(bc - 1)(ca - 1) = 0