Given that a and B satisfy b > a > e, where e is the base of natural logarithm, a ^ b > b ^ A is proved A ^ b > b ^ a B (LNA) > A (LNB) can you explain it?

A ^ b > b ^ a B (LNA) > A (LNB) (LNB) / be, so f (a) > F (b)
So (LNB) / BB ^ A
Add: take the logarithm ln (a ^ b) > ln (b ^ a) on both sides, and then we get that conclusion

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