不等式問題(求最少值)

50000 x (1.05^2n) - 50000 x (1.02^2n) > 64260 x (1.071^(n-1))
50000 x (1.05^2n) - 50000 x (1.02^2n) > 60000 x (1.071^n)
5 x (1.05^2n) - 5 x (1.02^2n) > 6 x (1.02 x 1.05)^n
5 x (1.05^2n) - 5 x (1.02^2n) > 6 x 1.02^n x 1.05^n
5 x (1.05^2n) - 6 x 1.02^n x 1.05^n - 5 x (1.02^2n) > 0
Let 1.05^n = x , 1.02^n = y

5x^2-6xy-5y^2>0
x / y < -0.566 or x / y > 1.766

1.05^n/1.02^n < -0.566 or 1.05^n/1.02^n > 1.766
(35/34)^n < -0.566(rejected) or (35/34)^n >1.766
n log (35/34) > log 1.766
n > 19.623

N的最少值為20

2014-04-10 14:16:28 補充：
(N係咪一定要整數)? 如果唔係一定整數 N最少值為19.623