📘 Core Rule to Compare \(a^b\) and \(b^a\)

Take natural logs on both sides. We compare \(a^b\) and \(b^a\) by comparing \(\;b\ln a\) and \(\;a\ln b\), i.e., \(\dfrac{\ln a}{a}\) and \(\dfrac{\ln b}{b}\):

\[ a^b \; \gtrless \; b^a \quad\Longleftrightarrow\quad \frac{\ln a}{a} \; \gtrless \; \frac{\ln b}{b}. \]

🎯 Two Instant Corollaries

1. Both bases \(>e\): If \(a>b>e\) then \(\dfrac{\ln a}{a}<\dfrac{\ln b}{b}\), hence \(\;a^bMnemonic: “When both bases exceed \(e\), the term with the bigger exponent wins.” (Because the smaller base goes to the bigger power.)
2. Both bases in \((0,e)\): If \(e>a>b>0\) then \(\dfrac{\ln a}{a}>\dfrac{\ln b}{b}\), hence \(\;a^b>b^a\).

🏔️ Peak of \(x^{1/x}\) occurs at \(x=e\)

Let \(f(x)=x^{1/x}=\exp\!\big(\tfrac{\ln x}{x}\big)\). Since \(\tfrac{\ln x}{x}\) peaks at \(x=e\), the maximum of \(x^{1/x}\) also occurs at \(x=e\) (global maximum for \(x>0\)).

\[ f'(x)=0 \;\Longleftrightarrow\; \frac{1-\ln x}{x^{2}}=0 \;\Longleftrightarrow\; x=e. \]

🧩 Application: Which is larger — \(1234^{4321}\) or \(4319^{1231}\)?

Step 1. Compare the swapped pair \(\,1234^{4321}\) vs \(\,4321^{1234}\). Since \(4321>1234>e\), by Corollary 1 \(\;1234^{4321} > 4321^{1234}\).\p>

Step 2. Now compare \(\,4321^{1234}\) and \(\,4319^{1234}\): larger base, same exponent \(\Rightarrow\;4321^{1234} > 4319^{1234}\).

Step 3. With the same base \(4319>e\), larger exponent wins: \(\,4319^{1234} > 4319^{1231}\).

Chain: \(\;1234^{4321} > 4321^{1234} > 4319^{1234} > 4319^{1231}\). Hence \( \boxed{\,1234^{4321} \;>\; 4319^{1231}\,} \).

⚡ Quick Checks

• \(11^{123}\) vs \(123^{11}\): both \(>e\), and \(11<123\) ⇒ \(11^{123} > 123^{11}\).
• \(2^{10}\) vs \(10^{2}\): here \(2e\). Use logs: compare \(10\ln2\) and \(2\ln10\): \(10\ln2\approx6.931 < 2\ln10\approx4.605?\) Actually \(10\ln2\approx6.93\) and \(2\ln10\approx4.61\) ⇒ \(2^{10} > 10^{2}\) (true).

📝 Quick Quiz