Exponent and Base Reversed

Question: Which of the two is larger, \(12^{11}\) or \(11^{12}\)

Solution

To solve any such questions, suppose we have two real numbers \(a\) and \(b\) that satisfy the inequality \(e \leq a \lt b\) where \(e\) is the natural exponent. Then, \(a^{b} > b^{a}\).

To prove this, consider changing the inequality slightly

\[ \begin{aligned} a^{b} &> b^{a}\newline \implies \left( a^{b} \right)^{1/ab} &> \left( b^{a} \right)^{1/ab}\newline \implies a^{1/a} &> b^{1/b} \end{aligned} \]

To show that the last equation is indeed true, consider the function \(x^{1/x}\) and its derivative

\[ \begin{aligned} y &= x^{1/x}\newline \ln y &= \frac{1}{x} \ln x\newline \frac{1}{y} \frac{dy}{dx} &= \frac{-1}{x^{2}} \ln x + \frac{1}{x} \frac{1}{x}\newline \implies \frac{dy}{dx} &= y \left( \frac{1}{x^{2}} (1 - \ln x) \right)\newline &= \frac{x^{1/x}}{x^{2}} \left( 1 - \ln x \right) \end{aligned} \]

The first term of the derivative is always positive for positive values of \(x\) (note: the function is only defined for positive values of \(x\)). As for the second term, its positive for \(x < e\) and vice versa. This means that \(x = e\) is the global maxima of the function as the derivative changes sign from positive to negative. Further, this is the only maxima for this function. Hence, for any two real numbers \(a\) and \(b\) satisfying \(e \leq a \lt b\), \(a^{1/a} > b^{1/b}\); which is what we set out to prove.

If both the numbers lie on the left side of \(e\), then the function is increasing and we reverse the inequality. For two numbers on the opposite sides of \(e\), directly making a claim is difficult and will require some algebra to derive the relation.