Let x and y be real numbers such that 3^x+13^y=17^x and 5^x+7^y=11^y, prove that x>y.
Thanks.
Real Mathematics
2014-09-04 1:33 am
回答 (2)
2014-09-04 9:08 pm
✔ 最佳答案
It should be x < y ?2014-09-04 13:08:54 補充：
If x ≥ y , then3ˣ + 13ʸ ≤ 3ˣ + 13ˣ
17ˣ ≤ 3ˣ + 13ˣ
1 ≤ (3/17)ˣ + (13/17)ˣ
It must be x < 1 since 1 > (3/17)¹ + (13/17)¹ = 16/17
and
5ˣ + 7ʸ ≥ 5ʸ + 7ʸ
11ʸ ≥ 5ʸ + 7ʸ
1 ≥ (5/11)ʸ + (7/11)ʸ
It must be y > 1 since 1 < (5/11)¹ + (7/11)¹ = 12/11 The above result gives x < 1 < y (Contradiction)
Therefore x < y.
In fact , x ≈ 1.05777 < y ≈ 1.1006 (Solved by Wolframalpha)
2014-09-04 8:59 pm
Yes, sorry, you are right, it should be x
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