指數題目，207^a = 243, 23^b = 9,求 (2a-5)(b+1).
[quote]207^a = 243 and 23^b = 9求 (2a-5)(b+1)[/quote]
207^a = 243 and 23^b = 9
→ 207^a = 3^5, 23^b=3^2
→ 207 = 3^(5/a), 23=3^(2/b)
兩式相除，3^2=3^{5/a - 2/b}
→ 5/a - 2/b = 2
→ 5b - 2a = 2ab
→ 2a + 2ab - 5b =0
→ (2a-5)(b+1) = - 5
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