$a^b > b^a$ or $a^b < b^a$ which is true if $a>b$ where $a,b$ are positive real numbers

Question

I was trying to find a simple relationship between mutual exponentiation of two positive real values, if one of them is greater/smaller to another.

user id:81360 131k · Accepted Answer

For positive $a,b$, note that $$ a^b > b^a \iff a^{1/a} > b^{1/b} $$ so effectively, you're asking whether the function $x^{1/x}$ is increasing or decreasing. As differentiation will show, this function is increasing for $0

The opposite will be true for $e

One example of it being insufficient is the special case $2^4=4^2$. — 2017-02-02

user id:82744 3k · Accepted Answer

Consider $a=3,b=2$

$2^3 \lt 3^2 \implies b^a \lt a^b$

Consider $a=4,b=-2$

$(-2)^4 \gt 4^{-2} \implies b^a \gt a^b$

So, it can be $b^a \lt a^b$ or $b^a \gt a^b$

$a^b > b^a$ or $a^b < b^a$ which is true if $a>b$ where $a,b$ are positive real numbers

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