Let $a , b > 2$ (with $a b \ne b$) ge e$be two numbers (for simplicity, let them be positive integers). The question is, which one is greater:$a^b$, or$b^a$? For instance, let$a=1000$and$b=999$. Which one is correct:$1000^{999} > 999^{1000}$, or$1000^{999} < 999^{1000}$? (I assume that equality does NOT hold.) Let the following hold:$|a-e| > |b-e|$. Then, it It can be shown that$a^b < b^a$. In fact, for every$x \ge 0$, we have$e^x > x^e$(assuming$x \ne e$). 4 added 17 characters in body I know a much simpler way; you just need to describe for him/her how to exponentiate. Let$a , b > 2$(with$a \ne b$) be two numbers (for simplicity, let them be positive integers). The question is, which one is greater:$a^b$, or$b^a$? For instance, let$a=1000$and$b=999$. Which one is correct:$1000^{999} > 999^{1000}$, or$1000^{999} < 999^{1000}$? (I assume that equality does NOT hold.) Let the following hold:$|a-e| > |b-e|$. Then, it can be shown that$a^b < b^a$. In fact, for every$x \ge 0$, we have$e^x > x^e$(assuming$x \ne e$). 3 edited body I know a much simpler way; you just need to describe for him/her how to exponentiate. Let$a , b > 1$2$ be two numbers (for simplicity, let them be positive integers). The question is, which one is greater: $a^b$, or $b^a$? For instance, let $a=1000$ and $b=999$. Which one is correct: $1000^{999} > 999^{1000}$, or $1000^{999} < 999^{1000}$? (I assume that equality does NOT hold.)
Let the following hold: $|a-e| > |b-e|$. Then, it can be shown that $a^b < b^a$.
In fact, for every $x \ge 0$, we have $e^x > x^e$ (assuming $x \ne e$).