帶指數既聯立方程.

(4^x) - (5^y) = 7……….式一
(4^x) = 7 + 5^y……….式一 b

[4^(x-1)] + [5^(y-2)] = 9……….式二
[4^(x) * 4^(-1)] + [5^(y) * 5^(-2)] = 9……….式二b
[4^(x) * 4^(-1)] * 4 * 5^2 + [5^(y) * 5^(-2)] * 4 * 5^2 = 9 * 4 * 5^2 ……….式二c
[4^(x) * 5^2] + [5^(y) * 4] = 9 * 4 * 5^2 ……….式二d

Substitute 式一 b into式二d, we get
[(7 + 5^y) * 5^2] + [5^(y) * 4] = 9 * 4 * 5^2
175 + 25(5^y) + 4(5^y) = 900
29 (5^y) = 900 - 175
29 (5^y) = 725
5^y = 25
5^y = 5^2
y = 2 (ans)

Substitute y = 2 into 式一, we get
(4^x) - (5^2) = 7
4^x - 25 = 7
4^x = 7 + 25 = 32 = 4^(2.5)
x = 2.5 (ans.)

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